Five-dimensional fermionic Chern-Simons theory
Dongsu Baka and Andreas Gustavssonb

a) Physics Department, University of Seoul, Seoul 02504, Korea
b) Department of Physics and Astronomy, Uppsala University,
Box 516, SE-75120 Uppsala, Sweden


(dsbak@uos.ac.kr, agbrev@gmail.com)


Abstract

We study 5d fermionic CS theory with a fermionic 2-form gauge potential. This theory can be obtained from 5d MSYM theory by performing the maximal topological twist. We put the theory on a five-manifold and compute the partition function. We find that it is a topological quantity, which involves the Ray-Singer torsion of the five-manifold. For abelian gauge group we consider the uplift to the 6d theory and find a mismatch between the 5d partition function and the 6d index, due to the nontrivial dimensional reduction of a selfdual two-form gauge field on a circle. We also discuss an application of the 5d theory to generalized knots made of 2d sheets embedded in 5d.

1 Introduction

Chern-Simons theory in 3d whose classical action is given by

k4πtr(AdA2i3AAA)𝑘4𝜋tr𝐴𝑑𝐴2𝑖3𝐴𝐴𝐴\displaystyle\frac{k}{4\pi}\int{\mbox{tr}}\left(A\wedge dA-\frac{2i}{3}A\wedge A\wedge A\right) (1.1)

has a long history. The seminal paper [1] obtained the exact result for the partition function for S3superscript𝑆3S^{3} by indirect methods. Later exact results have been obtained in [2, 3, 4] by various methods (nonabelian localization [5], abelianization, supersymmetric localization [6]) on a large class of three-manifolds. There have also been many works that have aimed to match such exact results with corresponding perturbative results in the large k𝑘k limit [7, 8, 9, 10, 11, 12]. Thus CS theory enables one to test path integral methods against known exact results.

We may generalize abelian CS theory to 2p+12𝑝12p+1 dimensions by taking the gauge potential to be a p𝑝p-form. When p𝑝p is odd, the gauge field is bosonic. However, when p𝑝p is even, a bosonic gauge field leads to a CS term that is a total derivative since ApdAp=12d(ApAp)subscript𝐴𝑝𝑑subscript𝐴𝑝12𝑑subscript𝐴𝑝subscript𝐴𝑝A_{p}\wedge dA_{p}=\frac{1}{2}d(A_{p}\wedge A_{p}). For even p𝑝p we shall therefore take the p𝑝p-form gauge field to be fermionic and then we have a fermionic CS theory or FCS theory for short. In the first few dimensions these CS and FCS actions, in Lorentzian signature and with canonical normalizations, are given by

S1dsubscript𝑆1𝑑\displaystyle S_{1d} =\displaystyle= i2ψ0dψ0𝑖2subscript𝜓0𝑑subscript𝜓0\displaystyle\frac{i}{2}\int\psi_{0}\wedge d\psi_{0} (1.2)
S3dsubscript𝑆3𝑑\displaystyle S_{3d} =\displaystyle= 12A1dA112subscript𝐴1𝑑subscript𝐴1\displaystyle\frac{1}{2}\int A_{1}\wedge dA_{1} (1.3)
S5dsubscript𝑆5𝑑\displaystyle S_{5d} =\displaystyle= i2ψ2dψ2𝑖2subscript𝜓2𝑑subscript𝜓2\displaystyle\frac{i}{2}\int\psi_{2}\wedge d\psi_{2} (1.4)
S7dsubscript𝑆7𝑑\displaystyle S_{7d} =\displaystyle= 12A3dA312subscript𝐴3𝑑subscript𝐴3\displaystyle\frac{1}{2}\int A_{3}\wedge dA_{3} (1.5)

The most general form of the gauge symmetry variations are

δψ0𝛿subscript𝜓0\displaystyle\delta\psi_{0} =\displaystyle= (χ0)0subscriptsubscript𝜒00\displaystyle(\chi_{0})_{0}
δA1𝛿subscript𝐴1\displaystyle\delta A_{1} =\displaystyle= dλ0+(λ1)0𝑑subscript𝜆0subscriptsubscript𝜆10\displaystyle d\lambda_{0}+(\lambda_{1})_{0}
δψ2𝛿subscript𝜓2\displaystyle\delta\psi_{2} =\displaystyle= dχ1+(χ2)0𝑑subscript𝜒1subscriptsubscript𝜒20\displaystyle d\chi_{1}+(\chi_{2})_{0}
δA3𝛿subscript𝐴3\displaystyle\delta A_{3} =\displaystyle= dλ2+(λ3)0𝑑subscript𝜆2subscriptsubscript𝜆30\displaystyle d\lambda_{2}+(\lambda_{3})_{0}

In addition to the usual exact forms, we shall also include the harmonic forms (χp)0subscriptsubscript𝜒𝑝0(\chi_{p})_{0} and (λp)0subscriptsubscript𝜆𝑝0(\lambda_{p})_{0} in order to have the most general closed forms by the Hodge decomposition [13]. The fact that abelian CS theories in various dimensions form a sequence (1.5) suggests that they could have some common features.

It is in general a quite difficult problem to generalize abelian higher rank gauge fields to nonabelian gauge groups. However, in 5d we automatically solve this problem since 5d FCS is obtained from 5d MSYM theory by performing the maximal twist. By this twist the SO(5)𝑆𝑂5SO(5) R-symmetry is identified with the SO(5)𝑆𝑂5SO(5) Lorentz symmetry [14, 15]. The twist gives one scalar nilpotent supercharge, which we can identify as the BRST charge associated with the two-form gauge symmetry, and the action can be interpreted as a BRST gauge fixed action for nonabelian 5d FCS theory.

For 3d CS on lens space S3/p=L(p;1)superscript𝑆3subscript𝑝𝐿𝑝1S^{3}/\mathbb{Z}_{p}=L(p;1), the exact partition function is known. From this exact result we can extract the perturbative expansion in 1/k1𝑘1/k. For gauge group G=SU(2)𝐺𝑆𝑈2G=SU(2), the resulting perturbative expansion for p𝑝p odd, is111If p𝑝p is even, then at =p/2𝑝2{{\ell}}=p/2 we get U𝑈U== diag(1,1)11(-1,-1) which commutes with all group elements in SU(2)𝑆𝑈2SU(2). In this case we should probably have the gauge group as SU(2)/2𝑆𝑈2subscript2SU(2)/\mathbb{Z}_{2} and identify this holonomy with the holonomy at =00{{\ell}}=0. We then do not count it since we only count gauge inequivalent holonomies. So for p𝑝p even, we sum over the holonomy sectors =0,1,,p/2101𝑝21{{\ell}}=0,1,...,p/2-1 and then this will again be in agreement with the general formula in [2].

Z𝑍\displaystyle Z =\displaystyle= eiπp4π21(iKp)3/2+eiπp422iKp=1p12e2πiK2p(sin2πp)2superscript𝑒𝑖𝜋𝑝4𝜋21superscript𝑖𝐾𝑝32superscript𝑒𝑖𝜋𝑝422𝑖𝐾𝑝superscriptsubscript1𝑝12superscript𝑒2𝜋𝑖𝐾superscript2𝑝superscript2𝜋𝑝2\displaystyle e^{\frac{i\pi p}{4}}\pi\sqrt{2}\frac{1}{\left(iKp\right)^{3/2}}+e^{\frac{i\pi p}{4}}\frac{2\sqrt{2}}{\sqrt{iKp}}\sum_{{{\ell}}=1}^{\frac{p-1}{2}}e^{\frac{2\pi iK{{\ell}}^{2}}{p}}\left(\sin\frac{2\pi{{\ell}}}{p}\right)^{2}

where K=k+2𝐾𝑘2K=k+2. This agrees with the perturbative expansion in [8]. We obtained this result from the exact result presented in [2] by expanding it out in powers of 1/k1𝑘1/k but where we suppress the next to leading orders in each sector labeled by =0,1,,p1201𝑝12{{\ell}}=0,1,...,\frac{p-1}{2}. This result can be rewritten in the form

Z𝑍\displaystyle Z =\displaystyle= eiπ4dim(G)ηgrav[1Vol(HA(0))τ0,SU(2)+=1p121Vol(HA())e2πiK2pτ,SU(2)]superscript𝑒𝑖𝜋4dimension𝐺subscript𝜂𝑔𝑟𝑎𝑣delimited-[]1Volsubscript𝐻superscript𝐴0subscript𝜏0𝑆𝑈2superscriptsubscript1𝑝121Volsubscript𝐻superscript𝐴superscript𝑒2𝜋𝑖𝐾superscript2𝑝subscript𝜏𝑆𝑈2\displaystyle e^{\frac{i\pi}{4}\dim(G)\eta_{grav}}\left[\frac{1}{{\mbox{Vol}}\left(H_{A^{(0)}}\right)}\sqrt{\tau_{0,SU(2)}}+\sum_{\ell=1}^{\frac{p-1}{2}}\frac{1}{{\mbox{Vol}}\left(H_{A^{(\ell)}}\right)}e^{\frac{2\pi iK\ell^{2}}{p}}\sqrt{\tau_{{{\ell}},SU(2)}}\right] (1.6)

Here HA()subscript𝐻superscript𝐴H_{A^{({{\ell}})}} denotes the unbroken gauge group by the gauge field background and τ,SU(2)subscript𝜏𝑆𝑈2\tau_{{{\ell}},SU(2)} denotes the Ray-Singer torsion of L(p;1)𝐿𝑝1L(p;1) associated with SU(2)𝑆𝑈2SU(2) gauge group and the holonomy labeled by =0,,p10𝑝1{{\ell}}=0,...,p-1. For the lens space L(p;q1,,qN1)=S2N1/p𝐿𝑝subscript𝑞1subscript𝑞𝑁1superscript𝑆2𝑁1subscript𝑝L(p;q_{1},\cdots,q_{N-1})=S^{2N-1}/\mathbb{Z}_{p} we have

τ0,SU(2)subscript𝜏0𝑆𝑈2\displaystyle\tau_{0,SU(2)} =\displaystyle= (τ0)3superscriptsubscript𝜏03\displaystyle(\tau_{0})^{3}
τ,SU(2)subscript𝜏𝑆𝑈2\displaystyle\tau_{{{\ell}},SU(2)} =\displaystyle= τ0τ2τ2(0)subscript𝜏0subscript𝜏2subscript𝜏20\displaystyle\tau_{0}\tau_{2{{\ell}}}\tau_{-2{{\ell}}}\qquad({{\ell}}\neq 0)

where

τ0subscript𝜏0\displaystyle\tau_{0} =\displaystyle= 1pN11superscript𝑝𝑁1\displaystyle\frac{1}{p^{N-1}}
τsubscript𝜏\displaystyle\tau_{{{\ell}}} =\displaystyle= |2Nsinπq11psinπqN11psinπp|superscript2𝑁𝜋superscriptsubscript𝑞11𝑝𝜋superscriptsubscript𝑞𝑁11𝑝𝜋𝑝\displaystyle\left|2^{N}\sin\frac{\pi q_{1}^{-1}{{\ell}}}{p}\cdots\sin\frac{\pi q_{N-1}^{-1}{{\ell}}}{p}\sin\frac{\pi{{\ell}}}{p}\right|

To get the torsion for L(p;1)𝐿𝑝1L(p;1), we shall put N=2𝑁2N=2 and q=q1=1𝑞superscript𝑞11q=q^{-1}=1. The unbroken gauge groups are HA(0)=SU(2)subscript𝐻superscript𝐴0𝑆𝑈2H_{A^{(0)}}=SU(2) and HA()=U(1)subscript𝐻superscript𝐴𝑈1H_{A^{({{\ell}})}}=U(1), whose volumes are

Vol(U(1))Vol𝑈1\displaystyle{\mbox{Vol}}(U(1)) =\displaystyle= 2πr2𝜋𝑟\displaystyle 2\pi r
Vol(SU(2))Vol𝑆𝑈2\displaystyle{\mbox{Vol}}(SU(2)) =\displaystyle= 2π2r32superscript𝜋2superscript𝑟3\displaystyle 2\pi^{2}r^{3}

We note that U(1)𝑈1U(1) corresponds to the equator of SU(2)=S3𝑆𝑈2superscript𝑆3SU(2)=S^{3}. The radius shall be chosen as

r𝑟\displaystyle r =\displaystyle= iπK2π𝑖𝜋𝐾2𝜋\displaystyle\sqrt{\frac{i}{\pi}}\sqrt{\frac{K}{2\pi}}

in order to match with the exact result. We can see why this value of the radius is natural up to the factor iπ𝑖𝜋\sqrt{\frac{i}{\pi}} as follows. We need to rescale A=K2πAcan𝐴𝐾2𝜋subscript𝐴𝑐𝑎𝑛A=\sqrt{\frac{K}{2\pi}}A_{can} to get a canonically normalized action. This can be achieved by rescaling the generators of SU(2)𝑆𝑈2SU(2) by the factor K2π𝐾2𝜋\sqrt{\frac{K}{2\pi}}. The overall factor eiπ4dim(G)ηgravsuperscript𝑒𝑖𝜋4dimension𝐺subscript𝜂𝑔𝑟𝑎𝑣e^{\frac{i\pi}{4}\dim(G)\eta_{grav}} is the remnant of the eta-invariant phase shift that results in the famous shift of the CS level from k𝑘k to K=k+2𝐾𝑘2K=k+2.

Next we consider 1d FCS on L(p)=S1/p𝐿𝑝superscript𝑆1subscript𝑝L(p)=S^{1}/\mathbb{Z}_{p} with gauge group SU(2)𝑆𝑈2SU(2) and the following action in Euclidean signature

S𝑆\displaystyle S =\displaystyle= k8π02π𝑑x0tr(ψD0ψ)𝑘8𝜋superscriptsubscript02𝜋differential-dsuperscript𝑥0tr𝜓subscript𝐷0𝜓\displaystyle\frac{k}{8\pi}\int_{0}^{2\pi}dx^{0}{\mbox{tr}}(\psi D_{0}\psi)

To compute the Witten index from the path integral, we do not need to use a Faddeev-Popov gauge fixing procedure since we can directly specify the gauge inequivalent gauge field configurations, which are classified by the holonomies

Pexpi02π𝑑x0A0(1001)=(e2πip00e2πip)𝑃𝑖superscriptsubscript02𝜋differential-dsuperscript𝑥0subscript𝐴0matrix1001matrixsuperscript𝑒2𝜋𝑖𝑝00superscript𝑒2𝜋𝑖𝑝\displaystyle P\exp i\int_{0}^{2\pi}dx^{0}A_{0}\begin{pmatrix}1&0\\ 0&-1\end{pmatrix}=\begin{pmatrix}e^{\frac{2\pi i{{\ell}}}{p}}&0\\ 0&e^{-\frac{2\pi i{{\ell}}}{p}}\end{pmatrix}

We can pick the gauge inequivalent gauge fields as

A0subscript𝐴0\displaystyle A_{0} =\displaystyle= p(1001)𝑝matrix1001\displaystyle\frac{{{\ell}}}{p}\begin{pmatrix}1&0\\ 0&-1\end{pmatrix}

and the path integral reduces to a discrete sum over {{\ell}}. This sum is presented in Eq. (3.1). But we can also carry out the standard Faddeev-Popov procedure and if we do that, then we are led to the result222As we will see, this result is correct only up to an overall phase factor.

I𝐼\displaystyle I =\displaystyle= 1Vol(HA(0))τ0,SU(2)+1Vol(HA())=1p12τ,SU(2)1Volsubscript𝐻superscript𝐴0subscript𝜏0𝑆𝑈21Volsubscript𝐻superscript𝐴superscriptsubscript1𝑝12subscript𝜏𝑆𝑈2\displaystyle\frac{1}{{\mbox{Vol}}(H_{A^{(0)}})}\sqrt{\tau_{0,SU(2)}}+\frac{1}{{\mbox{Vol}}(H_{A^{({{\ell}})}})}\sum_{{{\ell}}=1}^{\frac{p-1}{2}}\sqrt{\tau_{{{\ell}},SU(2)}}

where

τ0,SU(2)subscript𝜏0𝑆𝑈2\displaystyle\tau_{0,SU(2)} =\displaystyle= (τ0)3superscriptsubscript𝜏03\displaystyle(\tau_{0})^{3} (1.7)
τ,SU(2)subscript𝜏𝑆𝑈2\displaystyle\tau_{{{\ell}},SU(2)} =\displaystyle= τ0τ2τ2(0)subscript𝜏0subscript𝜏2subscript𝜏20\displaystyle\tau_{0}\tau_{2{{\ell}}}\tau_{-2{{\ell}}}\qquad({{\ell}}\neq 0) (1.8)

with

τ0subscript𝜏0\displaystyle\tau_{0} =\displaystyle= 11\displaystyle 1
τsubscript𝜏\displaystyle\tau_{{{\ell}}} =\displaystyle= |2sinπp|2𝜋𝑝\displaystyle\left|2\sin\frac{\pi{{\ell}}}{p}\right|

As we will show, the two expressions can be made to agree by taking the following radius for the SU(2)𝑆𝑈2SU(2) gauge group,

r𝑟\displaystyle r =\displaystyle= iπk2π𝑖𝜋𝑘2𝜋\displaystyle\sqrt{\frac{i}{\pi}}\sqrt{\frac{k}{2\pi}}

We notice that this radius takes the same form as we saw for 3d CS.

If we use our conjectured similarity between FCS theories in various dimensions, then we are led to the partition function

Z𝑍\displaystyle Z =\displaystyle= 1Vol(HA(0))τ0,SU(2)+=1p121Vol(HA())τ,SU(2)1Volsubscript𝐻superscript𝐴0subscript𝜏0𝑆𝑈2superscriptsubscript1𝑝121Volsubscript𝐻superscript𝐴subscript𝜏𝑆𝑈2\displaystyle\frac{1}{{\mbox{Vol}}\left(H_{A^{(0)}}\right)}\sqrt{\tau_{0,SU(2)}}+\sum_{\ell=1}^{\frac{p-1}{2}}\frac{1}{{\mbox{Vol}}\left(H_{A^{(\ell)}}\right)}\sqrt{\tau_{{{\ell}},SU(2)}} (1.9)

for 5d FCS with gauge group G=SU(2)𝐺𝑆𝑈2G=SU(2). We conjecture that the volume factors are on the same form as for 1d FCS when the 5d FCS action is canonically normalized, but we have not been able to explicitly compute the radius r𝑟r for this case. Gauge fixing amounts to adding BRST exact terms that we can also obtain by twisting of 5d MSYM. We will partly be able to confirm our conjecture by a localization computation in section 4.1.

We see that FCS theories differ from CS theories in many ways. Of course we do not know much about CS theories in other dimensions than three. For FCS, the partition function is one-loop exact. There is no phase factor multiplying the contributions from the various holonomy sectors, and there is no Chern-Simons level k𝑘k that can take arbitrary integer values for FCS.

The paper is organized as follows. In section 222 we construct explicit solutions for flat gauge fields on lens spaces in 3d and in 5d. In section 333 we compute the Witten index for 1d FCS. In section 444 we compute the partition function for 5d FCS. In section 555 we discuss applications to higher dimensional knots. In sections 666 and 777 we obtain the mismatch between the 5d partition function and the 6d Witten index that is related to the Ray-Singer torsion.

The appendices contain further details which makes the paper self-contained. In appendix A we review the definition and basic properties of the Ray-Singer torsion. In appendix A.1 we compute the Ray-Singer torsion on L(p;1,1)𝐿𝑝11L(p;1,1) in the trivial holonomy sector. In appendix B we present the Minakshisundaram-Pleijel theorem, which we use throughout the paper. In appendices C and D we address the problem of how to properly remove ghost zero modes. In appendix E we review what we need from 3d CS perturbation theory. In appendix F we present further details regarding the dimensional reduction on a circle and the mismatch related to the Ray-Singer torsion in various dimensions.

2 Flat gauge fields on lens spaces

For the lens space L2N1(g)=L~(p;q1,,qN)=S2N1/psubscript𝐿2𝑁1𝑔~𝐿𝑝subscript𝑞1subscript𝑞𝑁superscript𝑆2𝑁1subscript𝑝L_{2N-1}(g)=\widetilde{L}(p;q_{1},\dots,q_{N})=S^{2N-1}/\mathbb{Z}_{p}, the generator g𝑔g of psubscript𝑝\mathbb{Z}_{p} acts on Nsuperscript𝑁\mathbb{C}^{N} as

g(z1,,zN)=(e2πiq1/pz1,,e2πiqN/pzN)𝑔subscript𝑧1subscript𝑧𝑁superscript𝑒2𝜋𝑖subscript𝑞1𝑝subscript𝑧1superscript𝑒2𝜋𝑖subscript𝑞𝑁𝑝subscript𝑧𝑁\displaystyle g(z_{1},\dots,z_{N})=(e^{2\pi iq_{1}/p}z_{1},\dots,e^{2\pi iq_{N}/p}z_{N})

For twisted boundary conditions

gϕ𝑔italic-ϕ\displaystyle g\phi =\displaystyle= ω(g)ϕ𝜔𝑔italic-ϕ\displaystyle\omega(g)\phi

or explicitly

ϕ(e2πiq1/pz1,,e2πiqN/pzN)italic-ϕsuperscript𝑒2𝜋𝑖subscript𝑞1𝑝subscript𝑧1superscript𝑒2𝜋𝑖subscript𝑞𝑁𝑝subscript𝑧𝑁\displaystyle\phi(e^{2\pi iq_{1}/p}z_{1},\dots,e^{2\pi iq_{N}/p}z_{N}) =\displaystyle= e2πiq1/pϕ(z1,z2,z3)superscript𝑒2𝜋𝑖subscript𝑞1𝑝italic-ϕsubscript𝑧1subscript𝑧2subscript𝑧3\displaystyle e^{2\pi iq_{1}\ell/p}\phi(z_{1},z_{2},z_{3})

where we characterize ω𝜔\omega by the integer {{\ell}}, the Ray-Singer torsion was first computed by Ray by assuming N2𝑁2N\geq 2 in [16]. A direct computation has been made in [17, 18]. Lecture notes on the Ray-Singer torsion are [19, 20]. The result is

τω(L2N1(g))=|i=1N(2sinπqi1p)|subscript𝜏𝜔subscript𝐿2𝑁1𝑔superscriptsubscriptproduct𝑖1𝑁2𝜋superscriptsubscript𝑞𝑖1𝑝\displaystyle\tau_{\omega}(L_{2N-1}(g))=\left|\prod_{i=1}^{N}\left(2\sin\frac{\pi q_{i}^{-1}\ell}{p}\right)\right|

where we define qi1superscriptsubscript𝑞𝑖1q_{i}^{-1} as an integer such that

qi1qisuperscriptsubscript𝑞𝑖1subscript𝑞𝑖\displaystyle q_{i}^{-1}q_{i} \displaystyle\equiv 11\displaystyle 1 (2.1)

mod p𝑝p.333The absolute value seems unnatural. In Ray’s original computation [16], he computed the square of what we call the Ray-Singer torsion here. The square is real and positive and has no sign ambiguity under +p𝑝{{\ell}}\rightarrow{{\ell}}+p. In that sense it is the squared object that is the natural object to consider. But here we follow the widely used custom, and define the Ray-Singer torsion as the positive square root of Ray’s original definition of the torsion. When N𝑁N is even, the absolute value is not necessary. However, in this paper we will be considering both cases when N𝑁N is even and odd, so we need to have absolute value.

The condition (2.1) is also valid for N=1𝑁1N=1, which is a circle with a twisted boundary condition. We start with the lens space L~(p;1)~𝐿𝑝1\widetilde{L}(p;1), and twisted boundary condition

ϕ(e2πipz)italic-ϕsuperscript𝑒2𝜋𝑖𝑝𝑧\displaystyle\phi(e^{\frac{2\pi i}{p}}z) =\displaystyle= e2πipϕ(z)superscript𝑒2𝜋𝑖superscript𝑝italic-ϕ𝑧\displaystyle e^{\frac{2\pi i{{\ell}}^{\prime}}{p}}\phi(z) (2.2)

for some 0superscript0{{\ell}}^{\prime}\neq 0. By an explicit computation using the Hurwitz zeta function regularization, one can find the torsion

τ(L~(p;1))subscript𝜏superscript~𝐿𝑝1\displaystyle\tau_{{{\ell}}^{\prime}}(\widetilde{L}(p;1)) =\displaystyle= |2sinπp|2𝜋superscript𝑝\displaystyle\left|2\sin\frac{\pi{{\ell}}^{\prime}}{p}\right|

For the lens space L~(p;q1)~𝐿𝑝subscript𝑞1\widetilde{L}(p;q_{1}) we shall consider the twisted boundary condition

ϕ(e2πiq1pz)italic-ϕsuperscript𝑒2𝜋𝑖subscript𝑞1𝑝𝑧\displaystyle\phi(e^{\frac{2\pi iq_{1}}{p}}z) =\displaystyle= e2πipϕ(z)superscript𝑒2𝜋𝑖𝑝italic-ϕ𝑧\displaystyle e^{\frac{2\pi i{{\ell}}}{p}}\phi(z)

for some {{\ell}}. We get this boundary condition by iterating (2.2) q1subscript𝑞1q_{1} times, which gives =q1subscript𝑞1superscript{{\ell}}=q_{1}{{\ell}}^{\prime} and the torsion can be written in the form

τ(L~(p;q1))=|2sinπp|=|2sinπq11p|subscript𝜏~𝐿𝑝subscript𝑞12𝜋superscript𝑝2𝜋superscriptsubscript𝑞11𝑝\displaystyle\tau_{{{\ell}}}(\widetilde{L}(p;q_{1}))=\left|2\sin\frac{\pi{{\ell}}^{\prime}}{p}\right|=\left|2\sin\frac{\pi q_{1}^{-1}{{\ell}}}{p}\right|

This extends Ray’s computation, which is valid for N2𝑁2N\geq 2, to the case of N=1𝑁1N=1.

For a general lens space, we can always fix qN=1subscript𝑞𝑁1q_{N}=1 without imposing any restrictions. We then use the notation L(p;q1,qN1):=L~(p;q1,,qN1,1)assign𝐿𝑝subscript𝑞1subscript𝑞𝑁1~𝐿𝑝subscript𝑞1subscript𝑞𝑁11L(p;q_{1},\dots q_{N-1}):=\widetilde{L}(p;q_{1},\dots,q_{N-1},1) for the lens space.

We can also compute the Ray-Singer torsion with a trivial holonomy =00{{\ell}}=0. In this case the result is

τ0subscript𝜏0\displaystyle\tau_{0} =\displaystyle= 1pN11superscript𝑝𝑁1\displaystyle\frac{1}{p^{N-1}} (2.3)

We show this result by explicit computations in appendix A.1 for N=1𝑁1N=1 and N=3𝑁3N=3. The case of N=2𝑁2N=2 was addressed in the appendix of the paper [18]. From these results, we conjecture that the above formula will hold for all integers N=1,2,3,𝑁123N=1,2,3,....

Let us now obtain the Ray-Singer torsion for p𝑝p-forms taking values in the fundamental representation of the gauge group U(N)𝑈𝑁U(N). The lens space L(p;q1)𝐿𝑝subscript𝑞1L(p;q_{1}) has the fundamental group psubscript𝑝\mathbb{Z}_{p} and the holonomies are maps from the fundamental group into U(N)𝑈𝑁U(N), labeled by integers i=0,,p1subscript𝑖0𝑝1{{\ell}}_{i}=0,\dots,p-1 for i=1,,N𝑖1𝑁i=1,\dots,N. Alternatively we can consider a partition of N=N0+N1++Np1𝑁subscript𝑁0subscript𝑁1subscript𝑁𝑝1N=N_{0}+N_{1}+\cdots+N_{p-1} where Nsubscript𝑁N_{{{\ell}}}, for =0,,p10𝑝1{{\ell}}=0,\dots,p-1, counts the number of indices i=1,,N𝑖1𝑁i=1,\dots,N for which i=subscript𝑖{{\ell}}_{i}={{\ell}}. The twisted boundary conditions are such that the field component ϕijsubscriptitalic-ϕ𝑖𝑗\phi_{ij} carries charge ijsubscript𝑖subscript𝑗{{\ell}}_{i}-{{\ell}}_{j} under the U(1)𝑈1U(1) that rotates along the Hopf fiber of L(p;q1)𝐿𝑝subscript𝑞1L(p;q_{1}). The Ray-Singer torsion for U(N)𝑈𝑁U(N) is given by the product of the torsions for all the field components ij𝑖𝑗ij. This can be expressed as

τ(1,,N),U(N)=[τ0]Np1[τ]NNsubscript𝜏subscript1subscript𝑁𝑈𝑁superscriptdelimited-[]subscript𝜏0superscript𝑁superscriptsubscriptproductsuperscript𝑝1superscriptdelimited-[]subscript𝜏superscriptsubscript𝑁subscript𝑁superscript\displaystyle\tau_{({{\ell}}_{1},\dots,{{\ell}}_{N}),U(N)}=\left[\tau_{0}\right]^{N^{\prime}}\prod_{{{\ell}}\neq{{\ell}}^{\prime}}^{p-1}\left[\tau_{{{\ell}}-{{\ell}}^{\prime}}\right]^{N_{{{\ell}}}N_{{{\ell}}^{\prime}}}

where N==0p1N2Nsuperscript𝑁superscriptsubscript0𝑝1superscriptsubscript𝑁2𝑁N^{\prime}=\sum_{{{\ell}}=0}^{p-1}N_{{{\ell}}}^{2}\geq N. In the generic situation where all the isubscript𝑖{{\ell}}_{i}’s are distinct so that each N1subscript𝑁1N_{{{\ell}}}\leq 1, we have N=Nsuperscript𝑁𝑁N^{\prime}=N. To get the torsion for SU(N)𝑆𝑈𝑁SU(N) gauge group, we impose the restriction 1++N=0subscript1subscript𝑁0{{\ell}}_{1}+\cdots+{{\ell}}_{N}=0 mod p𝑝p. For SU(2)𝑆𝑈2SU(2) we get N=Np=1subscript𝑁subscript𝑁𝑝1N_{{{\ell}}}=N_{p-{{\ell}}}=1 for some {{\ell}}, and all other N=0subscript𝑁superscript0N_{{{\ell}}^{\prime}}=0. The Ray-Singer torsion then becomes

τ,SU(2)=τ0τ2τ2subscript𝜏𝑆𝑈2subscript𝜏0subscript𝜏2subscript𝜏2\displaystyle\tau_{{{\ell}},SU(2)}=\tau_{0}\tau_{2{{\ell}}}\tau_{-2{{\ell}}} =\displaystyle= 16p(sin2πq11psin2πp)216𝑝superscript2𝜋superscriptsubscript𝑞11𝑝2𝜋𝑝2\displaystyle\frac{16}{p}\left(\sin\frac{2\pi q_{1}^{-1}{{\ell}}}{p}\sin\frac{2\pi{{\ell}}}{p}\right)^{2}

We will now construct flat gauge fields with nontrivial holonomies for L(p;q)𝐿𝑝𝑞L(p;q) and for L(p;q1,q2)𝐿𝑝subscript𝑞1subscript𝑞2L(p;q_{1},q_{2}). We begin with considering the orbifold 2/psuperscript2subscript𝑝\mathbb{C}^{2}/\mathbb{Z}_{p} with the orbifold identification

(z1,z2)subscript𝑧1subscript𝑧2\displaystyle\left(z_{1},z_{2}\right) similar-to\displaystyle\sim (z1e2πiqp,z2e2πip)subscript𝑧1superscript𝑒2𝜋𝑖𝑞𝑝subscript𝑧2superscript𝑒2𝜋𝑖𝑝\displaystyle\left(z_{1}e^{\frac{2\pi iq}{p}},z_{2}e^{\frac{2\pi i}{p}}\right)

The lens space L(p;q)𝐿𝑝𝑞L(p;q) is defined by the equation |z1|2+|z2|2=1superscriptsubscript𝑧12superscriptsubscript𝑧221|z_{1}|^{2}+|z_{2}|^{2}=1 in this orbifold. But the lens space is smooth since the orbifold singularity is at the origin, away from the lens space. Let us consider two coordinate patches, U1={z10}={0<r}subscript𝑈1subscript𝑧100𝑟U_{1}=\{z_{1}\neq 0\}=\{0<r\} and U2={z20}={r<1}subscript𝑈2subscript𝑧20𝑟1U_{2}=\{z_{2}\neq 0\}=\{r<1\} with r=|z1|𝑟subscript𝑧1r=|z_{1}|. On the patch U2subscript𝑈2U_{2}, we define

z1superscript𝑧1\displaystyle z^{1} =\displaystyle= sinθ2eiqpψ2iϕ2𝜃2superscript𝑒𝑖𝑞𝑝subscript𝜓2𝑖subscriptitalic-ϕ2\displaystyle\sin\frac{\theta}{2}e^{i\frac{q}{p}\psi_{2}-i\phi_{2}}
z2superscript𝑧2\displaystyle z^{2} =\displaystyle= cosθ2ei1pψ2𝜃2superscript𝑒𝑖1𝑝subscript𝜓2\displaystyle\cos\frac{\theta}{2}e^{i\frac{1}{p}\psi_{2}}

We have the following rectangular T2superscript𝑇2T^{2}-identifications

ψ2subscript𝜓2\displaystyle\psi_{2} similar-to\displaystyle\sim ψ2+2πsubscript𝜓22𝜋\displaystyle\psi_{2}+2\pi
ϕ2subscriptitalic-ϕ2\displaystyle\phi_{2} similar-to\displaystyle\sim ϕ2+2πsubscriptitalic-ϕ22𝜋\displaystyle\phi_{2}+2\pi

Of course the metric on this T2superscript𝑇2T^{2} is complicated, induced from the flat metric on 2superscript2\mathbb{C}^{2}, but here the metric is not our concern. The above T2superscript𝑇2T^{2} identifications are preserved by the mapping class group SL(2,)𝑆𝐿2SL(2,\mathbb{Z}). To go to the patch U1subscript𝑈1U_{1} we shall preserve the T2superscript𝑇2T^{2} identifications, and hence the coordinate transformation must correspond to some element of SL(2,)𝑆𝐿2SL(2,\mathbb{Z}). Indeed this is the case. The transformation that does the job reads

ψ2subscript𝜓2\displaystyle\psi_{2} =\displaystyle= mψ1+pϕ1𝑚subscript𝜓1𝑝subscriptitalic-ϕ1\displaystyle m\psi_{1}+p\phi_{1}
ϕ2subscriptitalic-ϕ2\displaystyle\phi_{2} =\displaystyle= nψ1+qϕ1𝑛subscript𝜓1𝑞subscriptitalic-ϕ1\displaystyle n\psi_{1}+q\phi_{1}

where m𝑚m and n𝑛n are chosen so that

mqnp𝑚𝑞𝑛𝑝\displaystyle mq-np =\displaystyle= 11\displaystyle 1 (2.4)

The existence of such m𝑚m and n𝑛n follows from Bezout’s theorem and the assumption that p𝑝p and q𝑞q are relatively prime.

We have the same torus identifications after the SL(2,)𝑆𝐿2SL(2,\mathbb{Z}) transformation,

ψ1subscript𝜓1\displaystyle\psi_{1} similar-to\displaystyle\sim ψ1+2πsubscript𝜓12𝜋\displaystyle\psi_{1}+2\pi
ϕ1subscriptitalic-ϕ1\displaystyle\phi_{1} similar-to\displaystyle\sim ϕ1+2πsubscriptitalic-ϕ12𝜋\displaystyle\phi_{1}+2\pi

We get

z1superscript𝑧1\displaystyle z^{1} =\displaystyle= sinθ2ei1pψ1𝜃2superscript𝑒𝑖1𝑝subscript𝜓1\displaystyle\sin\frac{\theta}{2}e^{i\frac{1}{p}\psi_{1}}
z2superscript𝑧2\displaystyle z^{2} =\displaystyle= cosθ2eimpψ1+iϕ1𝜃2superscript𝑒𝑖𝑚𝑝subscript𝜓1𝑖subscriptitalic-ϕ1\displaystyle\cos\frac{\theta}{2}e^{i\frac{m}{p}\psi_{1}+i\phi_{1}}

which are coordinates on the patch U1subscript𝑈1U_{1}. In particular we notice that since mq1𝑚𝑞1mq\equiv 1 mod p𝑝p, we realize the lens space identification by taking ψ1ψ1+2πqsubscript𝜓1subscript𝜓12𝜋𝑞\psi_{1}\rightarrow\psi_{1}+2\pi q. Further since p𝑝p and q𝑞q are relative prime, we generate all elements of psubscript𝑝\mathbb{Z}_{p} by taking integer multiples of q𝑞q if we count modulo p𝑝p.

We now seek a flat connection which corresponds to the holonomy

exp2πip2𝜋𝑖𝑝\displaystyle\exp\frac{2\pi i\ell}{p}

when integrated over the closed path C𝐶C that is specified by (for t[0,2π]𝑡02𝜋t\in[0,2\pi])

(ψ1ϕ1)matrixsubscript𝜓1subscriptitalic-ϕ1\displaystyle\begin{pmatrix}\psi_{1}\\ \phi_{1}\end{pmatrix} =\displaystyle= (t0)matrix𝑡0\displaystyle\begin{pmatrix}t\\ 0\end{pmatrix}

using the coordinates on U1subscript𝑈1U_{1}, and by

(ψ2ϕ2)matrixsubscript𝜓2subscriptitalic-ϕ2\displaystyle\begin{pmatrix}\psi_{2}\\ \phi_{2}\end{pmatrix} =\displaystyle= (mtnt)matrix𝑚𝑡𝑛𝑡\displaystyle\begin{pmatrix}mt\\ nt\end{pmatrix}

using the coordinates on U2subscript𝑈2U_{2}.

One such flat connection on the patch U1subscript𝑈1U_{1} is given by

A|U1evaluated-at𝐴subscript𝑈1\displaystyle A|_{U_{1}} =\displaystyle= pdψ1𝑝𝑑subscript𝜓1\displaystyle\frac{\ell}{p}d\psi_{1}

which is well defined over the entire U1subscript𝑈1U_{1}. Note that over U1subscript𝑈1U_{1}, the T2superscript𝑇2T^{2} collapses to a circle when |z2|=0subscript𝑧20|z_{2}|=0 or r=1𝑟1r=1 and, even in the region near the circle, this connection A1subscript𝐴1A_{1} is well defined. In the overlap region U1U2subscript𝑈1subscript𝑈2U_{1}\cap U_{2}, the coordinate transformed version of A|U1evaluated-at𝐴subscript𝑈1A|_{U_{1}} is given by

A|U1U2evaluated-at𝐴subscript𝑈1subscript𝑈2\displaystyle A|_{U_{1}\cap U_{2}} =\displaystyle= p(qdψ2pdϕ2)𝑝𝑞𝑑subscript𝜓2𝑝𝑑subscriptitalic-ϕ2\displaystyle\frac{\ell}{p}\left(qd\psi_{2}-pd\phi_{2}\right)

Again note that over U2subscript𝑈2U_{2}, T2superscript𝑇2T^{2} collapses to a circle when |z1|=r=0subscript𝑧1𝑟0|z_{1}|=r=0. To make A𝐴A well defined all over U2subscript𝑈2U_{2}, we need to make a large gauge transformation

A=A+dϕ2=qpdψ2superscript𝐴𝐴𝑑subscriptitalic-ϕ2𝑞𝑝𝑑subscript𝜓2\displaystyle A^{\prime}=A+\ell d\phi_{2}=\frac{q\ell}{p}d\psi_{2}

in the overlap of U1U2subscript𝑈1subscript𝑈2U_{1}\cap U_{2}. Then the connection

A|U2evaluated-atsuperscript𝐴subscript𝑈2\displaystyle A^{\prime}|_{U_{2}} =\displaystyle= qpdψ2𝑞𝑝𝑑subscript𝜓2\displaystyle\frac{q{{\ell}}}{p}d\psi_{2}

is well-defined all over U2subscript𝑈2U_{2}.

Using (2.4) it is easy to see that

expiCA=expiCA=exp2πip𝑖subscript𝐶𝐴𝑖subscript𝐶superscript𝐴2𝜋𝑖𝑝\displaystyle\exp i\int_{C}A=\exp i\int_{C}A^{\prime}=\exp\frac{2\pi i{{\ell}}}{p}

if C𝐶C is in U1U2subscript𝑈1subscript𝑈2U_{1}\cap U_{2}. If C𝐶C is not in U1subscript𝑈1U_{1}, then we compute the holonomy using the expression expiCA𝑖subscript𝐶superscript𝐴\exp i\int_{C}A^{\prime} and if C𝐶C is not in U2subscript𝑈2U_{2}, then we use expiCA𝑖subscript𝐶𝐴\exp i\int_{C}A and we get the same result. Thus we have shown that there exists a flat gauge field on L(p;q)𝐿𝑝𝑞L(p;q) with the above given holonomy around C𝐶C.

In the fundamental representation of SU(2)𝑆𝑈2SU(2) gauge group we wish to find a flat gauge field corresponding to the holonomy

Usubscript𝑈\displaystyle U_{\ell} =\displaystyle= (ω00ωp)matrixsuperscript𝜔00superscript𝜔𝑝\displaystyle\begin{pmatrix}\omega^{\ell}&0\\ 0&\omega^{p-\ell}\end{pmatrix}

where we define ω=e2πi/p𝜔superscript𝑒2𝜋𝑖𝑝\omega=e^{2\pi i/p}. In the path integral, we shall sum over all gauge inequivalent holonomies. Since the Weyl group, which is a subgroup of the gauge group, permutes the two diagonal elements in the holonomy, we see that gauge inequivalent holonomies are obtained by restricting the range to =0,1,,[p/2]01delimited-[]𝑝2\ell=0,1,...,[p/2] where [p/2]=p/2delimited-[]𝑝2𝑝2[p/2]=p/2 if p𝑝p is even, and (p1)/2𝑝12(p-1)/2 if p𝑝p is odd. If p𝑝p is even, then we find two elements in the center, U0=diag(1,1)subscript𝑈0𝑑𝑖𝑎𝑔11U_{0}=diag(1,1) and Up/2=(1,1)subscript𝑈𝑝211U_{p/2}=(-1,-1). If p𝑝p is odd, the only element in the center is U0subscript𝑈0U_{0}.

A corresponding flat gauge field in the fundamental representation of SU(2)𝑆𝑈2SU(2) is given by

A|U1evaluated-at𝐴subscript𝑈1\displaystyle A|_{U_{1}} =\displaystyle= p(1001)dψ1𝑝matrix1001𝑑subscript𝜓1\displaystyle\frac{{{\ell}}}{p}\begin{pmatrix}1&0\\ 0&-1\end{pmatrix}d\psi_{1}

For U(1)𝑈1U(1) gauge group, flat gauge fields on the lens space L(p;q)𝐿𝑝𝑞L(p;q) are classified by the holonomy around the Hopf fiber

expiCA𝑖subscript𝐶𝐴\displaystyle\exp i\int_{C}A =\displaystyle= exp2πip2𝜋𝑖𝑝\displaystyle\exp\frac{2\pi i\ell}{p}

where =0,,p10𝑝1\ell=0,...,p-1. These take values in π1(L(p;q))=psubscript𝜋1𝐿𝑝𝑞subscript𝑝\pi_{1}(L(p;q))=\mathbb{Z}_{p}. The Chern-Simons action is given by

S𝑆\displaystyle S =\displaystyle= ik4πL(p;q)AdA𝑖𝑘4𝜋subscript𝐿𝑝𝑞𝐴𝑑𝐴\displaystyle-\frac{ik}{4\pi}\int_{L(p;q)}A\wedge dA

in Euclidean signature. This action has been explicitly computed for q=p1𝑞𝑝1q=p-1 on a flat gauge field in [21]. The result is444Since we are in Euclidean signature, we need an i𝑖i in the CS action. The minus sign in the right hand side is related to the choice of orientation of the lens space.

exp(S)𝑆\displaystyle\exp(-S) =\displaystyle= exp(πik1pp2)𝜋𝑖𝑘1𝑝𝑝superscript2\displaystyle\exp\left(-\pi ik\frac{1-p}{p}\ell^{2}\right)

This result is nontrivial, despite F=dA=0𝐹𝑑𝐴0F=dA=0 and so the CS action would naively be zero. But the above definition of the CS action is not entirely correct when we need to cover the manifold with many patches and the gauge field is related by gauge transformations as we pass from one patch to another. A better way to define the CS action in such a situation, is as SM4FFsimilar-to𝑆subscriptsubscript𝑀4𝐹𝐹S\sim\int_{M_{4}}F\wedge F where M4=L(p;q)subscript𝑀4𝐿𝑝𝑞\partial M_{4}=L(p;q). Here F𝐹F does not have to vanish on M4subscript𝑀4M_{4}. For the detailed construction of F𝐹F, we refer to [21].

For SU(2)𝑆𝑈2SU(2) gauge group the value of Chern-Simons action has been obtained for a flat gauge field corresponding to the holonomy

PexpiCA𝑃𝑖subscript𝐶𝐴\displaystyle P\exp i\int_{C}A =\displaystyle= (exp2πip00exp2πip)matrix2𝜋𝑖𝑝002𝜋𝑖𝑝\displaystyle\begin{pmatrix}\exp\frac{2\pi i\ell}{p}&0\\ 0&\exp-\frac{2\pi i\ell}{p}\end{pmatrix}

for all lens spaces in [22]. The result is

exp(S)𝑆\displaystyle\exp(-S) =\displaystyle= exp(2πikq12p)2𝜋𝑖𝑘superscript𝑞1superscript2𝑝\displaystyle\exp\left(\frac{2\pi ikq^{-1}\ell^{2}}{p}\right)

where q1q=1superscript𝑞1𝑞1q^{-1}q=1 mod p𝑝p. For q=p1𝑞𝑝1q=p-1 we have q1=p1superscript𝑞1𝑝1q^{-1}=p-1 and

exp(S)=exp(2πikp1p2)=exp(2πik2p)𝑆2𝜋𝑖𝑘𝑝1𝑝superscript22𝜋𝑖𝑘superscript2𝑝\displaystyle\exp(-S)=\exp\left(2\pi ik\frac{p-1}{p}\ell^{2}\right)=\exp\left(-\frac{2\pi ik\ell^{2}}{p}\right)

There is a factor 222 compared to the U(1)𝑈1U(1) case, which we can understand as follows. The SU(2)𝑆𝑈2SU(2) CS action on this flat gauge field with the above holonomy is given by

AdA+(A)d(A)=2AdA𝐴𝑑𝐴𝐴𝑑𝐴2𝐴𝑑𝐴\displaystyle AdA+(-A)d(-A)=2AdA

where A𝐴A denotes the U(1)𝑈1U(1) gauge field. We may also notice that the minus sign in the second step is consistent with taking k𝑘k to k𝑘-k in the resulting CS action for the lens space L(p;1)𝐿𝑝1L(p;1). The lens spaces L(p;1)𝐿𝑝1L(p;1) and L(p;p1)𝐿𝑝𝑝1L(p;p-1) are related by parity, which flips the sign of k𝑘k.

We now move on the L(p;q1,q2)𝐿𝑝subscript𝑞1subscript𝑞2L(p;q_{1},q_{2}). We view S5superscript𝑆5S^{5} as a circle bundle over 2superscript2\mathbb{CP}^{2} that we can cover with three patches

Uasubscript𝑈𝑎\displaystyle U_{a} =\displaystyle= {(z1,z2,z3)S53|za0}conditional-setsubscript𝑧1subscript𝑧2subscript𝑧3superscript𝑆5superscript3subscript𝑧𝑎0\displaystyle\{(z_{1},z_{2},z_{3})\in S^{5}\subset\mathbb{C}^{3}|z_{a}\neq 0\}

The lens space L(p;q1,q2)𝐿𝑝subscript𝑞1subscript𝑞2L(p;q_{1},q_{2}) is defined as S5/psuperscript𝑆5subscript𝑝S^{5}/\mathbb{Z}_{p} where

(z1,z2,z3)subscript𝑧1subscript𝑧2subscript𝑧3\displaystyle\left(z_{1},z_{2},z_{3}\right) similar-to\displaystyle\sim (z1e2πiq1p,z2e2πiq2p,z3e2πip)subscript𝑧1superscript𝑒2𝜋𝑖subscript𝑞1𝑝subscript𝑧2superscript𝑒2𝜋𝑖subscript𝑞2𝑝subscript𝑧3superscript𝑒2𝜋𝑖𝑝\displaystyle\left(z_{1}e^{\frac{2\pi iq_{1}}{p}},z_{2}e^{\frac{2\pi iq_{2}}{p}},z_{3}e^{\frac{2\pi i}{p}}\right) (2.5)

On U3subscript𝑈3U_{3} we use the coordinates

z1subscript𝑧1\displaystyle z_{1} =\displaystyle= sinχcosθ2eiq1py3+iψ3+iϕ3𝜒𝜃2superscript𝑒𝑖subscript𝑞1𝑝subscript𝑦3𝑖subscript𝜓3𝑖subscriptitalic-ϕ3\displaystyle\sin\chi\cos\frac{\theta}{2}e^{i\frac{q_{1}}{p}y_{3}+i\psi_{3}+i\phi_{3}}
z2subscript𝑧2\displaystyle z_{2} =\displaystyle= sinχsinθ2eiq2py3iψ3𝜒𝜃2superscript𝑒𝑖subscript𝑞2𝑝subscript𝑦3𝑖subscript𝜓3\displaystyle\sin\chi\sin\frac{\theta}{2}e^{i\frac{q_{2}}{p}y_{3}-i\psi_{3}}
z3subscript𝑧3\displaystyle z_{3} =\displaystyle= cosχei1py3𝜒superscript𝑒𝑖1𝑝subscript𝑦3\displaystyle\cos\chi e^{i\frac{1}{p}y_{3}}

The lens space identification (2.5) is obtained by taking y3y3+2πsubscript𝑦3subscript𝑦32𝜋y_{3}\rightarrow y_{3}+2\pi. The coordinates χ𝜒\chi and θ𝜃\theta may be used on all three patches if we understand that their ranges are different depending on the patch,

U3subscript𝑈3\displaystyle U_{3} =\displaystyle= {0χ<π/2,0θπ}formulae-sequence0𝜒𝜋20𝜃𝜋\displaystyle\{0\leq\chi<\pi/2,\quad 0\leq\theta\leq\pi\}
U2subscript𝑈2\displaystyle U_{2} =\displaystyle= {0<χπ/2,0<θπ}formulae-sequence0𝜒𝜋20𝜃𝜋\displaystyle\{0<\chi\leq\pi/2,\quad 0<\theta\leq\pi\}
U1subscript𝑈1\displaystyle U_{1} =\displaystyle= {0<χπ/2,0θ<π}formulae-sequence0𝜒𝜋20𝜃𝜋\displaystyle\{0<\chi\leq\pi/2,\quad 0\leq\theta<\pi\}

The coordinates (y3,ψ3,ϕ3)subscript𝑦3subscript𝜓3subscriptitalic-ϕ3(y_{3},\psi_{3},\phi_{3}) take values in a three-torus T3=3/(2π)3superscript𝑇3superscript3superscript2𝜋3T^{3}=\mathbb{R}^{3}/(2\pi\mathbb{Z})^{3}. We map from U3subscript𝑈3U_{3} to U2subscript𝑈2U_{2} by the following SL(3,)𝑆𝐿3SL(3,\mathbb{Z}) coordinate transformation

(y3ψ3ϕ3)matrixsubscript𝑦3subscript𝜓3subscriptitalic-ϕ3\displaystyle\begin{pmatrix}y_{3}\\ \psi_{3}\\ \phi_{3}\end{pmatrix} =\displaystyle= (mp0nq20n1q1q21)(y2ψ2ϕ2)matrix𝑚𝑝0𝑛subscript𝑞20𝑛1subscript𝑞1subscript𝑞21matrixsubscript𝑦2subscript𝜓2subscriptitalic-ϕ2\displaystyle\begin{pmatrix}m&p&0\\ n&q_{2}&0\\ -n&1-q_{1}-q_{2}&1\end{pmatrix}\begin{pmatrix}y_{2}\\ \psi_{2}\\ \phi_{2}\end{pmatrix}

where m𝑚m and n𝑛n are such that

mq2np𝑚subscript𝑞2𝑛𝑝\displaystyle mq_{2}-np =\displaystyle= 11\displaystyle 1 (2.6)

Since SL(3,)𝑆𝐿3SL(3,\mathbb{Z}) is the mapping class group of T3superscript𝑇3T^{3}, we have (y2,ϕ2,ϕ2)T3subscript𝑦2subscriptitalic-ϕ2subscriptitalic-ϕ2superscript𝑇3(y_{2},\phi_{2},\phi_{2})\in T^{3} and

z1subscript𝑧1\displaystyle z_{1} =\displaystyle= sinχcosθ2eiq1mpy2+iψ2+iϕ2𝜒𝜃2superscript𝑒𝑖subscript𝑞1𝑚𝑝subscript𝑦2𝑖subscript𝜓2𝑖subscriptitalic-ϕ2\displaystyle\sin\chi\cos\frac{\theta}{2}e^{i\frac{q_{1}m}{p}y_{2}+i\psi_{2}+i\phi_{2}}
z2subscript𝑧2\displaystyle z_{2} =\displaystyle= sinχsinθ2ei1py2𝜒𝜃2superscript𝑒𝑖1𝑝subscript𝑦2\displaystyle\sin\chi\sin\frac{\theta}{2}e^{i\frac{1}{p}y_{2}}
z3subscript𝑧3\displaystyle z_{3} =\displaystyle= cosχeimpy2+iψ2𝜒superscript𝑒𝑖𝑚𝑝subscript𝑦2𝑖subscript𝜓2\displaystyle\cos\chi e^{i\frac{m}{p}y_{2}+i\psi_{2}}

and the lens space identification (2.5) is obtained by taking y2y2+2πq2subscript𝑦2subscript𝑦22𝜋subscript𝑞2y_{2}\rightarrow y_{2}+2\pi q_{2}.

We map from U3subscript𝑈3U_{3} to U1subscript𝑈1U_{1} by the following SL(3,)𝑆𝐿3SL(3,\mathbb{Z}) transformation,

(y3ψ3ϕ3)matrixsubscript𝑦3subscript𝜓3subscriptitalic-ϕ3\displaystyle\begin{pmatrix}y_{3}\\ \psi_{3}\\ \phi_{3}\end{pmatrix} =\displaystyle= (m~pp01q2q2n~1+q1+q2q1+q2)(y1ψ1ϕ1)matrix~𝑚𝑝𝑝01subscript𝑞2subscript𝑞2~𝑛1subscript𝑞1subscript𝑞2subscript𝑞1subscript𝑞2matrixsubscript𝑦1subscript𝜓1subscriptitalic-ϕ1\displaystyle\begin{pmatrix}\widetilde{m}&-p&-p\\ 0&1-q_{2}&-q_{2}\\ -\widetilde{n}&-1+q_{1}+q_{2}&q_{1}+q_{2}\end{pmatrix}\begin{pmatrix}y_{1}\\ \psi_{1}\\ \phi_{1}\end{pmatrix}

where m~~𝑚\widetilde{m} and n~~𝑛\widetilde{n} are such that

m~q1n~p~𝑚subscript𝑞1~𝑛𝑝\displaystyle\widetilde{m}q_{1}-\widetilde{n}p =\displaystyle= 11\displaystyle 1 (2.7)

Then we get

z1subscript𝑧1\displaystyle z_{1} =\displaystyle= sinχcosθ2ei1py1𝜒𝜃2superscript𝑒𝑖1𝑝subscript𝑦1\displaystyle\sin\chi\cos\frac{\theta}{2}e^{i\frac{1}{p}y_{1}}
z2subscript𝑧2\displaystyle z_{2} =\displaystyle= sinχsinθ2eiq2m~py1iψ1𝜒𝜃2superscript𝑒𝑖subscript𝑞2~𝑚𝑝subscript𝑦1𝑖subscript𝜓1\displaystyle\sin\chi\sin\frac{\theta}{2}e^{i\frac{q_{2}\widetilde{m}}{p}y_{1}-i\psi_{1}}
z3subscript𝑧3\displaystyle z_{3} =\displaystyle= cosχeim~py1iψ1iϕ1𝜒superscript𝑒𝑖~𝑚𝑝subscript𝑦1𝑖subscript𝜓1𝑖subscriptitalic-ϕ1\displaystyle\cos\chi e^{i\frac{\widetilde{m}}{p}y_{1}-i\psi_{1}-i\phi_{1}}

By using q1m~1subscript𝑞1~𝑚1q_{1}\widetilde{m}\equiv 1 mod p𝑝p, we find the lens space identification (2.5) by taking y1y1+2πq1subscript𝑦1subscript𝑦12𝜋subscript𝑞1y_{1}\rightarrow y_{1}+2\pi q_{1}.

The map from U2subscript𝑈2U_{2} to U1subscript𝑈1U_{1} can be obtained by composing the map from U2subscript𝑈2U_{2} to U3subscript𝑈3U_{3} (the inverse of the map from U3subscript𝑈3U_{3} to U2subscript𝑈2U_{2}) and the map from U3subscript𝑈3U_{3} to U1subscript𝑈1U_{1}.

In the overlap U1U2U3subscript𝑈1subscript𝑈2subscript𝑈3U_{1}\cap U_{2}\cap U_{3}, we take the gauge field as

A𝐴\displaystyle A =\displaystyle= pdy3=p(mdy2+pdψ2)=p(m~dy1pdψ1pdϕ1)𝑝𝑑subscript𝑦3𝑝𝑚𝑑subscript𝑦2𝑝𝑑subscript𝜓2𝑝~𝑚𝑑subscript𝑦1𝑝𝑑subscript𝜓1𝑝𝑑subscriptitalic-ϕ1\displaystyle\frac{\ell}{p}dy_{3}=\frac{\ell}{p}\left(mdy_{2}+pd\psi_{2}\right)=\frac{\ell}{p}\left(\widetilde{m}dy_{1}-pd\psi_{1}-pd\phi_{1}\right)

Just as we did in 3d case, here again we shall remove Dirac string singularities in order to have a well-defined gauge potential on the patches U2subscript𝑈2U_{2} and U3subscript𝑈3U_{3}. Thus we define

A|U3evaluated-at𝐴subscript𝑈3\displaystyle A|_{U_{3}} =\displaystyle= pdy3𝑝𝑑subscript𝑦3\displaystyle\frac{{{\ell}}}{p}dy_{3}
A|U2evaluated-atsuperscript𝐴subscript𝑈2\displaystyle A^{\prime}|_{U_{2}} =\displaystyle= pmdy2𝑝𝑚𝑑subscript𝑦2\displaystyle\frac{{{\ell}}}{p}mdy_{2}
A′′|U1evaluated-atsuperscript𝐴′′subscript𝑈1\displaystyle A^{\prime\prime}|_{U_{1}} =\displaystyle= pm~dy1𝑝~𝑚𝑑subscript𝑦1\displaystyle\frac{{{\ell}}}{p}\widetilde{m}dy_{1}

On the overlap regions these gauge fields are related by large gauge transformations. The path C𝐶C along which we integrate the holonomy is expressed as follows in the three coordinate patches respectively as follows,

(y3,ψ3,ϕ3)subscript𝑦3subscript𝜓3subscriptitalic-ϕ3\displaystyle(y_{3},\psi_{3},\phi_{3}) =\displaystyle= (1,0,0)t100𝑡\displaystyle(1,0,0)t
(y2,ψ2,ϕ2)subscript𝑦2subscript𝜓2subscriptitalic-ϕ2\displaystyle(y_{2},\psi_{2},\phi_{2}) =\displaystyle= (q2,n,(1q1)n)tsubscript𝑞2𝑛1subscript𝑞1𝑛𝑡\displaystyle(q_{2},-n,(1-q_{1})n)t
(y1,ψ1,ϕ1)subscript𝑦1subscript𝜓1subscriptitalic-ϕ1\displaystyle(y_{1},\psi_{1},\phi_{1}) =\displaystyle= (q1,q2n~,(1q2)n~)tsubscript𝑞1subscript𝑞2~𝑛1subscript𝑞2~𝑛𝑡\displaystyle(q_{1},q_{2}\widetilde{n},(1-q_{2})\widetilde{n})t

where t[0,2π]𝑡02𝜋t\in[0,2\pi]. The holonomy remains the same after the large gauge transformations and is the same irrespectively of which patch we compute it in and is given by

expiCA=expiCA=expiCA′′=exp2πip𝑖subscript𝐶𝐴𝑖subscript𝐶superscript𝐴𝑖subscript𝐶superscript𝐴′′2𝜋𝑖𝑝\displaystyle\exp i\int_{C}A=\exp i\int_{C}A^{\prime}=\exp i\int_{C}A^{\prime\prime}=\exp\frac{2\pi i{{\ell}}}{p}

To see this, we use the relations (2.6) and (2.7).

For SU(2)𝑆𝑈2SU(2) gauge group we also have a flat gauge field that we get simply multiplying the U(1)𝑈1U(1) flat gauge field by the matrix diag(1,1)11(1,-1) and for U(N)𝑈𝑁U(N) gauge group we have the flat gauge field

A|U3evaluated-at𝐴subscript𝑈3\displaystyle A|_{U_{3}} =\displaystyle= 1p(1N)dy31𝑝matrixsubscript1missing-subexpressionmissing-subexpressionmissing-subexpressionmissing-subexpressionmissing-subexpressionmissing-subexpressionsubscript𝑁𝑑subscript𝑦3\displaystyle\frac{1}{p}\begin{pmatrix}{{\ell}}_{1}&&\\ &\ddots&\\ &&{{\ell}}_{N}\end{pmatrix}dy_{3}

Thus we find that to each possible holonomy

exp2πip(1N)2𝜋𝑖𝑝matrixsubscript1missing-subexpressionmissing-subexpressionmissing-subexpressionmissing-subexpressionmissing-subexpressionmissing-subexpressionsubscript𝑁\displaystyle\exp\frac{2\pi i}{p}\begin{pmatrix}{{\ell}}_{1}&&\\ &\ddots&\\ &&{{\ell}}_{N}\end{pmatrix}

around the fiber of L(p;q1,q2)𝐿𝑝subscript𝑞1subscript𝑞2L(p;q_{1},q_{2}), there is a corresponding flat gauge field, which is defined on each coordinate patch and related to different patches by gauge transformations.

3 One-dimensional fermionic Chern-Simons

Before turning to the more complicated case of 5d FCS, we will consider 1d FCS, that is, quantum mechanics with one real fermion. Let us consider a hermitian operator a𝑎a subject to the algebra

a2superscript𝑎2\displaystyle a^{2} =\displaystyle= 𝟙1\displaystyle\mathds{1}

and assume that there is one state |0ket0\left|0\right>. Acting with a𝑎a we get another state |1=a|0ket1𝑎ket0\left|1\right>=a\left|0\right>. It is not possible for a𝑎a to annihilate |0ket0\left|0\right>, since by acting twice by a𝑎a we shall get back |0ket0\left|0\right>. If we act by a𝑎a on |1ket1\left|1\right> we get a|1=|0𝑎ket1ket0a\left|1\right>=\left|0\right> by using a2=1superscript𝑎21a^{2}=1. Let us normalize the state as 0|0=1inner-product001\Braket{0}{0}=1. Inserting 𝟙=aa1superscript𝑎𝑎\mathds{1}=a^{{\dagger}}a, we get 1|1=1inner-product111\Braket{1}{1}=1. We have 0|1=1|0=0inner-product01inner-product100\Braket{0}{1}=\Braket{1}{0}=0 as a consequence of the requirement that the 2×2222\times 2 matrix realization of a𝑎a shall square to the identity matrix. We have the completeness relation

|00|+|11|ket0bra0ket1bra1\displaystyle\left|0\right>\left<0\right|+\left|1\right>\left<1\right| =\displaystyle= 𝟙1\displaystyle\mathds{1}

We introduce a Grassmann odd parameter ψ𝜓\psi such that ψψ=0𝜓𝜓0\psi\psi=0 which we can integrate over with the usual rules

𝑑ψdifferential-d𝜓\displaystyle\int d\psi =\displaystyle= 00\displaystyle 0
𝑑ψψdifferential-d𝜓𝜓\displaystyle\int d\psi\psi =\displaystyle= 11\displaystyle 1

We have the anticommutativity property,

aψ𝑎𝜓\displaystyle a\psi =\displaystyle= ψa𝜓𝑎\displaystyle-\psi a

We define the state

|ψket𝜓\displaystyle\left|\psi\right> =\displaystyle= |0+|1ψket0ket1𝜓\displaystyle\left|0\right>+\left|1\right>\psi

The conjugate state is

ψ|bra𝜓\displaystyle\left<\psi\right| =\displaystyle= 0|+ψ1|bra0𝜓bra1\displaystyle\left<0\right|+\psi\left<1\right|

We have the following properties

ψ|ψinner-productsuperscript𝜓𝜓\displaystyle\Braket{\psi^{\prime}}{\psi} =\displaystyle= eψψsuperscript𝑒superscript𝜓𝜓\displaystyle e^{\psi^{\prime}\psi}
ψ|a|ψquantum-operator-productsuperscript𝜓𝑎𝜓\displaystyle\Braket{\psi^{\prime}}{a}{\psi} =\displaystyle= ψ+ψsuperscript𝜓𝜓\displaystyle\psi^{\prime}+\psi
𝑑ψ|ψψ|adifferential-d𝜓ket𝜓bra𝜓𝑎\displaystyle\int d\psi\left|-\psi\right>\left<\psi\right|a =\displaystyle= 𝟙1\displaystyle\mathds{1}
𝑑ψψ|aA|ψdifferential-d𝜓quantum-operator-product𝜓𝑎𝐴𝜓\displaystyle\int d\psi\left<\psi\right|aA\left|\psi\right> =\displaystyle= trAtr𝐴\displaystyle{\mbox{tr}}A

The partition function can be written as

Z𝑍\displaystyle Z =\displaystyle= tr𝟙=𝑑ψψ|a|ψ=𝑑ψψ|a(aa)(aa)(aa)|ψtr1differential-d𝜓quantum-operator-product𝜓𝑎𝜓differential-d𝜓quantum-operator-product𝜓𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝜓\displaystyle{\mbox{tr}}\mathds{1}=\int d\psi\left<\psi\right|a\left|\psi\right>=\int d\psi\left<\psi\right|a(aa)(aa)\cdots(aa)\left|\psi\right>
=\displaystyle= 𝑑ψψ|𝟙a(𝟙a𝟙a)(𝟙a𝟙a)|ψdifferential-d𝜓quantum-operator-product𝜓1𝑎1𝑎1𝑎1𝑎1𝑎𝜓\displaystyle\int d\psi\left<\psi\right|\mathds{1}a(\mathds{1}a\mathds{1}a)\cdots(\mathds{1}a\mathds{1}a)\left|\psi\right>

We use the completeness relation which absorbs all the operators a𝑎a and we get

Z𝑍\displaystyle Z =\displaystyle= 𝑑ψ𝑑ψN𝑑ψ1ψ|ψNψN|ψN1ψ1|ψdifferential-d𝜓differential-dsubscript𝜓𝑁differential-dsubscript𝜓1inner-product𝜓subscript𝜓𝑁inner-productsubscript𝜓𝑁subscript𝜓𝑁1inner-productsubscript𝜓1𝜓\displaystyle\int d\psi d\psi_{N}\cdots d\psi_{1}\Braket{\psi}{-\psi_{N}}\Braket{\psi_{N}}{-\psi_{N-1}}\cdots\Braket{\psi_{1}}{\psi}
=\displaystyle= 𝑑ψ𝑑ψN𝑑ψ1eψψN+ψNψN1++ψ1ψdifferential-d𝜓differential-dsubscript𝜓𝑁differential-dsubscript𝜓1superscript𝑒𝜓subscript𝜓𝑁subscript𝜓𝑁subscript𝜓𝑁1subscript𝜓1𝜓\displaystyle\int d\psi d\psi_{N}\cdots d\psi_{1}e^{-\psi\psi_{N}+\psi_{N}\psi_{N-1}+\cdots+\psi_{1}\psi}
=\displaystyle= 𝑑ψ𝑑ψN𝑑ψ1e(ϵ(ψ)(ψ)ψNϵ++ϵψ1ψ1ψϵ)differential-d𝜓differential-dsubscript𝜓𝑁differential-dsubscript𝜓1superscript𝑒italic-ϵ𝜓𝜓subscript𝜓𝑁italic-ϵitalic-ϵsubscript𝜓1subscript𝜓1𝜓italic-ϵ\displaystyle\int d\psi d\psi_{N}\cdots d\psi_{1}e^{-\left(\epsilon(-\psi)\frac{(-\psi)-\psi_{N}}{\epsilon}+\cdots+\epsilon\psi_{1}\frac{\psi_{1}-\psi}{\epsilon}\right)}

We can compute this partition function for any odd integer N𝑁N and always get the same answer,

Z𝑍\displaystyle Z =\displaystyle= 22\displaystyle 2

which counts the number of states. In the limit N𝑁N\rightarrow\infty, we get the path integral

Z𝑍\displaystyle Z =\displaystyle= 𝒟ψeiS𝒟𝜓superscript𝑒𝑖𝑆\displaystyle\int{\cal{D}}\psi e^{iS}

where we take the antiperiodic boundary condition ψ(2π)=ψ(0)𝜓2𝜋𝜓0\psi(2\pi)=-\psi(0) and the FCS action is given by

S𝑆\displaystyle S =\displaystyle= k4π02π𝑑tψψ˙𝑘4𝜋superscriptsubscript02𝜋differential-d𝑡𝜓˙𝜓\displaystyle\frac{k}{4\pi}\int_{0}^{2\pi}dt\psi\dot{\psi}

Here k𝑘k is a real parameter that will be determined by matching with canonical quantization. Since the fermionic fields are anticommuting, the FCS action is purely imaginary. We find that this choice is necessary in order for the partition function to become a real number. We compute the path integral by expanding the field in orthonormalized modes with respect to the metric (or line element) dt𝑑𝑡dt. Our mode expansion reads

ψ(t)𝜓𝑡\displaystyle\psi(t) =\displaystyle= nψn+12ei(n+12)tsubscript𝑛subscript𝜓𝑛12superscript𝑒𝑖𝑛12𝑡\displaystyle\sum_{n\in\mathbb{Z}}\psi_{n+\frac{1}{2}}e^{i\left(n+\frac{1}{2}\right)t}

The action is

S𝑆\displaystyle S =\displaystyle= ikn=0(n+12)ψn+12ψn12𝑖𝑘superscriptsubscript𝑛0𝑛12subscript𝜓𝑛12subscript𝜓𝑛12\displaystyle-ik\sum_{n=0}^{\infty}\left(n+\frac{1}{2}\right)\psi_{n+\frac{1}{2}}\psi_{-n-\frac{1}{2}}

We compute this using Hurwitz zeta function as follows,

Z𝑍\displaystyle Z =\displaystyle= n=0[k(n+12)]=keζ(0,12)=2ksuperscriptsubscriptproduct𝑛0delimited-[]𝑘𝑛12𝑘superscript𝑒superscript𝜁0122𝑘\displaystyle\prod_{n=0}^{\infty}\left[k\left(n+\frac{1}{2}\right)\right]=\sqrt{k}e^{-\zeta^{\prime}(0,\frac{1}{2})}=\sqrt{2k}

where we choose the fermionic measure as

𝒟ψ=n=0dψn+12dψn12𝒟𝜓superscriptsubscriptproduct𝑛0𝑑subscript𝜓𝑛12𝑑subscript𝜓𝑛12\displaystyle{\cal{D}}\psi=\prod_{n=0}^{\infty}d\psi_{n+\frac{1}{2}}d\psi_{-n-\frac{1}{2}}

To match with the result from canonical quantization, that is Z=2𝑍2Z=2, we shall take k=2𝑘2k=2.

Let us next compute the Witten index. We define a fermion number operator as

(1)F|0superscript1𝐹ket0\displaystyle(-1)^{F}\left|0\right> =\displaystyle= |0ket0\displaystyle\left|0\right>
(1)F|1superscript1𝐹ket1\displaystyle(-1)^{F}\left|1\right> =\displaystyle= |1ket1\displaystyle-\left|1\right>

This means that

(1)Fasuperscript1𝐹𝑎\displaystyle(-1)^{F}a =\displaystyle= a(1)F𝑎superscript1𝐹\displaystyle-a(-1)^{F}
(1)F|ψsuperscript1𝐹ket𝜓\displaystyle(-1)^{F}\left|\psi\right> =\displaystyle= |ψket𝜓\displaystyle\left|-\psi\right>

The Witten index is

I=tr(1)F=𝑑ψψ|a(1)F|ψ=𝑑ψψ|a|ψ𝐼trsuperscript1𝐹differential-d𝜓quantum-operator-product𝜓𝑎superscript1𝐹𝜓differential-d𝜓quantum-operator-product𝜓𝑎𝜓\displaystyle I={\mbox{tr}}(-1)^{F}=\int d\psi\left<\psi\right|a(-1)^{F}\left|\psi\right>=-\int d\psi\left<-\psi\right|a\left|\psi\right>

If we then follow through the same steps as above, we end up with the same path integral but with the boundary condition ψ(2π)=ψ(0)𝜓2𝜋𝜓0\psi(2\pi)=\psi(0) and in that case we get

I𝐼\displaystyle I =\displaystyle= 00\displaystyle 0

when we compute the resulting expression for any odd number N𝑁N of steps. This is easy to see for N=1𝑁1N=1. Then the exponent becomes ψψ1+ψ1ψ=0𝜓subscript𝜓1subscript𝜓1𝜓0\psi\psi_{1}+\psi_{1}\psi=0 and the integration over ψ𝜓\psi and ψ1subscript𝜓1\psi_{1} yields zero.

Since a2=𝟙superscript𝑎21a^{2}=\mathds{1}, we can introduce projection operators

P±subscript𝑃plus-or-minus\displaystyle P_{\pm} =\displaystyle= 12(1±a)12plus-or-minus1𝑎\displaystyle\frac{1}{2}\left(1\pm a\right)

Orthonormalized eigenstates of P±subscript𝑃plus-or-minusP_{\pm} are

|±ketplus-or-minus\displaystyle\left|\pm\right> =\displaystyle= 12(|0±|1)12plus-or-minusket0ket1\displaystyle\frac{1}{\sqrt{2}}\left(\left|0\right>\pm\left|1\right>\right)

Now a|±=±|±𝑎ketplus-or-minusplus-or-minusketplus-or-minusa\left|\pm\right>=\pm\left|\pm\right>, and (1)Fsuperscript1𝐹(-1)^{F} maps the two states into each other, (1)F|±=|superscript1𝐹ketplus-or-minusketminus-or-plus(-1)^{F}\left|\pm\right>=\left|\mp\right>. We have the projected Witten index

I+=tr((1)FP+)=+|(1)F|+=+|=0subscript𝐼trsuperscript1𝐹subscript𝑃delimited-⟨⟩limit-fromsuperscript1𝐹inner-product0\displaystyle I_{+}={\mbox{tr}}\left((-1)^{F}P_{+}\right)=\left<+\right|(-1)^{F}\left|+\right>=\Braket{+}{-}=0

We can imagine a different definition where we instead define (1)F=asuperscript1𝐹𝑎(-1)^{F}=a. With this definition we declare |+ket\left|+\right> is bosonic and |ket\left|-\right> is fermionic and then we get

I+=tr((1)FP+)=+|a|+=+|+=1subscript𝐼trsuperscript1𝐹subscript𝑃delimited-⟨⟩limit-from𝑎inner-product1\displaystyle I_{+}={\mbox{tr}}\left((-1)^{F}P_{+}\right)=\left<+\right|a\left|+\right>=\Braket{+}{+}=1

To connect with the path integral, an attempt would be to define

|ψket𝜓\displaystyle\left|\psi\right> =\displaystyle= eψ|+superscript𝑒𝜓ket\displaystyle e^{\psi}\left|+\right>

since then, by using aψ=ψa𝑎𝜓𝜓𝑎a\psi=-\psi a,

(1)F|ψsuperscript1𝐹ket𝜓\displaystyle(-1)^{F}\left|\psi\right> =\displaystyle= |ψket𝜓\displaystyle\left|-\psi\right>

which leads to a path integral with periodic boundary condition. However, the action turns out to become nonlocal.

Let us next compute the Witten index by taking out the zero mode. We then consider the path integral

I𝐼\displaystyle I =\displaystyle= 𝒟ψeiS𝒟𝜓superscript𝑒𝑖𝑆\displaystyle\int{\cal{D}}\psi e^{iS}

with periodic boundary condition. We expand the field in modes

ψ(t)𝜓𝑡\displaystyle\psi(t) =\displaystyle= n0ψneintsubscript𝑛0subscript𝜓𝑛superscript𝑒𝑖𝑛𝑡\displaystyle\sum_{n\neq 0}\psi_{n}e^{int}

The action becomes

S𝑆\displaystyle S =\displaystyle= ikn=1nψnψn𝑖𝑘superscriptsubscript𝑛1𝑛subscript𝜓𝑛subscript𝜓𝑛\displaystyle-ik\sum_{n=1}^{\infty}n\psi_{n}\psi_{-n}

The index becomes

Iosc=n=1kn=2πksubscript𝐼𝑜𝑠𝑐superscriptsubscriptproduct𝑛1𝑘𝑛2𝜋𝑘\displaystyle I_{osc}=\prod_{n=1}^{\infty}kn=\sqrt{\frac{2\pi}{k}}

where we use the fermionic measure

𝒟ψ=n=1dψndψn𝒟𝜓superscriptsubscriptproduct𝑛1𝑑subscript𝜓𝑛𝑑subscript𝜓𝑛\displaystyle{\cal{D}}\psi=\prod_{n=1}^{\infty}d\psi_{n}d\psi_{-n}

Now we have gauge fixed a fermionic zero mode. As we illustrate in the appendix C, when we do this, we shall also divide the result by the volume of the gauge group. Here the gauge group depends on the context. If the gauge group is trivial, we shall not divide by anything. But if the gauge group is U(1)𝑈1U(1) acting trivially on ψ𝜓\psi in the adjoint representation, then we shall divide by its volume V𝑉V.

Let us now consider the following nonabelian generalization with the action

S𝑆\displaystyle S =\displaystyle= k8π02π𝑑ttrR(ψDtψ)𝑘8𝜋superscriptsubscript02𝜋differential-d𝑡subscripttr𝑅𝜓subscript𝐷𝑡𝜓\displaystyle\frac{k}{8\pi}\int_{0}^{2\pi}dt{\mbox{tr}}_{R}(\psi D_{t}\psi)

where the trace is taken in some representation R𝑅R of the gauge group. We define Dtψ=ψ˙i[At,ψ]subscript𝐷𝑡𝜓˙𝜓𝑖subscript𝐴𝑡𝜓D_{t}\psi=\dot{\psi}-i[A_{t},\psi] where Atsubscript𝐴𝑡A_{t} is a background gauge field. Let us now restrict to the gauge group SU(2)𝑆𝑈2SU(2) and let the gauge field be At=At3T3subscript𝐴𝑡subscript𝐴𝑡3superscript𝑇3A_{t}=A_{t3}T^{3} where [T3,T±]=±T±superscript𝑇3superscript𝑇plus-or-minusplus-or-minussuperscript𝑇plus-or-minus[T^{3},T^{\pm}]=\pm T^{\pm}. We expand ψ=ψ3T3+ψ+T++ψT𝜓subscript𝜓3superscript𝑇3subscript𝜓superscript𝑇subscript𝜓superscript𝑇\psi=\psi_{3}T^{3}+\psi_{+}T^{+}+\psi_{-}T^{-} and find

Dtψ3subscript𝐷𝑡subscript𝜓3\displaystyle D_{t}\psi_{3} =\displaystyle= ψ˙3subscript˙𝜓3\displaystyle\dot{\psi}_{3}
Dtψ±subscript𝐷𝑡subscript𝜓plus-or-minus\displaystyle D_{t}\psi_{\pm} =\displaystyle= ψ˙±iAt3ψ±minus-or-plussubscript˙𝜓plus-or-minus𝑖subscript𝐴𝑡3subscript𝜓plus-or-minus\displaystyle\dot{\psi}_{\pm}\mp iA_{t3}\psi_{\pm}

Let us assume the generators are in the fundamental representation, R=𝑅R=\Box, of SU(2)𝑆𝑈2SU(2) and let us normalize the generators so that tr(T3T3)=2subscripttrsuperscript𝑇3superscript𝑇32{\mbox{tr}}_{\Box}(T^{3}T^{3})=2 and tr(T+T)=1subscripttrsuperscript𝑇superscript𝑇1{\mbox{tr}}_{\Box}(T^{+}T^{-})=1. Then the action becomes a sum of two terms, S=S3+S+𝑆subscript𝑆3subscript𝑆S=S_{3}+S_{+} where

S3subscript𝑆3\displaystyle S_{3} =\displaystyle= k4π02π𝑑tψ3ψ˙3𝑘4𝜋superscriptsubscript02𝜋differential-d𝑡subscript𝜓3subscript˙𝜓3\displaystyle\frac{k}{4\pi}\int_{0}^{2\pi}dt\psi_{3}\dot{\psi}_{3}
S+subscript𝑆\displaystyle S_{+} =\displaystyle= k4π02π𝑑t(ψ+ψ˙iAt3ψ+ψ)𝑘4𝜋superscriptsubscript02𝜋differential-d𝑡subscript𝜓subscript˙𝜓𝑖subscript𝐴𝑡3subscript𝜓subscript𝜓\displaystyle\frac{k}{4\pi}\int_{0}^{2\pi}dt\left(\psi_{+}\dot{\psi}_{-}-iA_{t3}\psi_{+}\psi_{-}\right)

The holonomy is

U𝑈\displaystyle U =\displaystyle= (eih00eih)matrixsuperscript𝑒𝑖00superscript𝑒𝑖\displaystyle\begin{pmatrix}e^{ih}&0\\ 0&e^{-ih}\end{pmatrix}

where h=02π𝑑tAt3superscriptsubscript02𝜋differential-d𝑡subscript𝐴𝑡3h=\int_{0}^{2\pi}dtA_{t3}. The Witten index splits into a product I=I3I+𝐼subscript𝐼3subscript𝐼I=I_{3}I_{+}. From the previous result we get from S3subscript𝑆3S_{3} the contribution I3=Iosc/Vsubscript𝐼3subscript𝐼𝑜𝑠𝑐𝑉I_{3}=I_{osc}/V. From S+subscript𝑆S_{+} we get

I+=n(n+h2π)=2isin(h2)subscript𝐼subscriptproduct𝑛𝑛2𝜋2𝑖2\displaystyle I_{+}=\prod_{n\in\mathbb{Z}}\left(n+\frac{h}{2\pi}\right)=2i\sin\left(\frac{h}{2}\right)

for h00h\neq 0, where in the last step we used zeta function regularization. This does not depend on k𝑘k. We also know that the Witten index shall be ±2isin(h/2)plus-or-minus2𝑖2\pm 2i\sin(h/2) from canonical quantization, since this is the sum over two states, spin-down and spin-down, with a relative minus sign as we are computing a Witten index. This gives the sine function multiplied by the factor of ±2iplus-or-minus2𝑖\pm 2i. The sign depends on how we define the fermion number F𝐹F when we define the Witten index – whether the spin-up state is taken as the fermionic or the bosonic state.

For general gauge group, holonomy U𝟙𝑈1U\neq\mathds{1} in the representation R𝑅R, the Ray-Singer torsion on S1superscript𝑆1S^{1} is given by [23, 24]

τ𝜏\displaystyle\tau =\displaystyle= det(𝟙U)R\displaystyle\det{}_{R}(\mathds{1}-U)

If we specialize this formula to the fundamental representation of SU(2)𝑆𝑈2SU(2), then this yields

τ=(1eih)(1eih)=(2sin(h2))2𝜏1superscript𝑒𝑖1superscript𝑒𝑖superscript222\displaystyle\tau=(1-e^{ih})(1-e^{-ih})=\left(2\sin\left(\frac{h}{2}\right)\right)^{2}

We see that up to a phase factor ±iplus-or-minus𝑖\pm i, the fermionic Chern-Simons action computes the square root of the RS torsion on S1superscript𝑆1S^{1}.

We can restrict ourselves to h=2π(2)/p2𝜋2𝑝h=2\pi(2{{\ell}})/p in which case we can view this as the torsion of the lens space L(p)=S1/p𝐿𝑝superscript𝑆1subscript𝑝L(p)=S^{1}/\mathbb{Z}_{p}. The partition function is given by

I𝐼\displaystyle I =\displaystyle= I3I+(0)+=1p12I3I+()subscript𝐼3subscript𝐼0superscriptsubscript1𝑝12subscript𝐼3subscript𝐼\displaystyle I_{3}I_{+}(0)+\sum_{{{\ell}}=1}^{\frac{p-1}{2}}I_{3}I_{+}({{\ell}}) (3.1)

where

I3subscript𝐼3\displaystyle I_{3} =\displaystyle= IoscVsubscript𝐼𝑜𝑠𝑐𝑉\displaystyle\frac{I_{osc}}{V}
I+(0)subscript𝐼0\displaystyle I_{+}(0) =\displaystyle= (Iosc)2superscriptsubscript𝐼𝑜𝑠𝑐2\displaystyle(I_{osc})^{2}
I+()subscript𝐼\displaystyle I_{+}({{\ell}}) =\displaystyle= 2isin2πp2𝑖2𝜋𝑝\displaystyle 2i\sin\frac{2\pi{{\ell}}}{p}

This expression is explained as follows: when =00{{\ell}}=0 we get the zero mode contribution from ψ3,ψ+,ψsubscript𝜓3subscript𝜓subscript𝜓\psi_{3},\psi_{+},\psi_{-}, whereas when 00{{\ell}}\neq 0 we get the zero mode contribution only from ψ3subscript𝜓3\psi_{3}. We note that U(1)𝑈1U(1) inside SU(2)𝑆𝑈2SU(2) is generated by T3superscript𝑇3T^{3}, so for I3subscript𝐼3I_{3} we divide by V𝑉V but for I+(0)subscript𝐼0I_{+}(0) we do not divide by V𝑉V since there is no U(1)𝑈1U(1) associated with I+(0)subscript𝐼0I_{+}(0). We get the squared expression I+=(Iosc)2subscript𝐼superscriptsubscript𝐼𝑜𝑠𝑐2I_{+}=(I_{osc})^{2} as can be understood by rewriting the Lagrangian in terms of real spinor components: ψ+ψ˙=ψ1ψ˙1+ψ2ψ˙2subscript𝜓subscript˙𝜓subscript𝜓1subscript˙𝜓1subscript𝜓2subscript˙𝜓2\psi_{+}\dot{\psi}_{-}=\psi_{1}\dot{\psi}_{1}+\psi_{2}\dot{\psi}_{2}. Now we can write this in the form

I𝐼\displaystyle I =\displaystyle= e3πi2[1Vol(SU(2))τ0,SU(2)+1Vol(U(1))=1p12τ,SU(2)]superscript𝑒3𝜋𝑖2delimited-[]1Vol𝑆𝑈2subscript𝜏0𝑆𝑈21Vol𝑈1superscriptsubscript1𝑝12subscript𝜏𝑆𝑈2\displaystyle e^{\frac{3\pi i}{2}}\left[\frac{1}{{\mbox{Vol}}(SU(2))}\sqrt{\tau_{0,SU(2)}}+\frac{1}{{\mbox{Vol}}(U(1))}\sum_{{{\ell}}=1}^{\frac{p-1}{2}}\sqrt{\tau_{{{\ell}},SU(2)}}\right] (3.2)

where the radius of SU(2)𝑆𝑈2SU(2) and the volume V𝑉V shall be taken as

r𝑟\displaystyle r =\displaystyle= iπk2π,𝑖𝜋𝑘2𝜋\displaystyle\sqrt{\frac{i}{\pi}}\sqrt{\frac{k}{2\pi}}, (3.3)
V𝑉\displaystyle V =\displaystyle= 2π2𝜋\displaystyle 2\sqrt{\pi} (3.4)

To show this result, we need to notice that I+()subscript𝐼I_{+}({{\ell}}) lies in the upper halfplane for all values =1,,(p1)/21𝑝12{{\ell}}=1,...,(p-1)/2. The overall phase factor e3πi2superscript𝑒3𝜋𝑖2e^{\frac{3\pi i}{2}} is should somehow come from the BRST gauge fixing of the fermionic zero mode, since that gives the same factor for all the holonomy sectors. We notice that by requiring V𝑉V coincides with Vol(U(1))=2πrVol𝑈12𝜋𝑟{\mbox{Vol}}(U(1))=2\pi r we have to fix k=2πi𝑘2𝜋𝑖k=-2\pi i. For this value the abelian FCS action becomes

S𝑆\displaystyle S =\displaystyle= i202π𝑑tψψ˙𝑖2superscriptsubscript02𝜋differential-d𝑡𝜓˙𝜓\displaystyle-\frac{i}{2}\int_{0}^{2\pi}dt\psi\dot{\psi}

which is both real and canonically normalized.

4 Five-dimensional fermionic Chern-Simons

In [14] we found that maximally twisted 5d MSYM gives a 5d fermionic Chern-Simons theory. This is a topological field theory. We are now interested in its partition function. Since the action is topological, the partition function should be a topological invariant of the five-manifold M5subscript𝑀5M_{5}.

Let us begin with assuming the gauge group is U(1)𝑈1U(1). If we introduce the linear combinations

𝒜msubscript𝒜𝑚\displaystyle{\cal{A}}_{m} =\displaystyle= Amϕmsubscript𝐴𝑚subscriptitalic-ϕ𝑚\displaystyle A_{m}-\phi_{m}
𝒜¯msubscript¯𝒜𝑚\displaystyle\overline{\cal{A}}_{m} =\displaystyle= Am+ϕmsubscript𝐴𝑚subscriptitalic-ϕ𝑚\displaystyle A_{m}+\phi_{m}

then the Lagrangian can be expressed as

=FCS+δVsubscript𝐹𝐶𝑆𝛿𝑉\displaystyle{\cal{L}}={\cal{L}}_{FCS}+\delta V

where the fermionic Chern-Simons term is

FCSsubscript𝐹𝐶𝑆\displaystyle{\cal{L}}_{FCS} =\displaystyle= i8ϵmnpqrψmnpψqr𝑖8superscriptitalic-ϵ𝑚𝑛𝑝𝑞𝑟subscript𝜓𝑚𝑛subscript𝑝subscript𝜓𝑞𝑟\displaystyle-\frac{i}{8}\epsilon^{mnpqr}\psi_{mn}\partial_{p}\psi_{qr}

and the gauge fixing fermion that arises from twisting of 5d MSYM is given by

V𝑉\displaystyle V =\displaystyle= 14¯mnψmn12ϕψϕmmψ14subscript¯𝑚𝑛superscript𝜓𝑚𝑛12italic-ϕ𝜓superscriptitalic-ϕ𝑚subscript𝑚𝜓\displaystyle\frac{1}{4}\overline{{\cal{F}}}_{mn}\psi^{mn}-\frac{1}{2}\phi\psi-\phi^{m}\partial_{m}\psi

By an integration by parts we can write this as

V𝑉\displaystyle V =\displaystyle= 12𝒜¯n𝒟¯mψmn12ϕψϕmmψ12subscript¯𝒜𝑛subscript¯𝒟𝑚superscript𝜓𝑚𝑛12italic-ϕ𝜓superscriptitalic-ϕ𝑚subscript𝑚𝜓\displaystyle-\frac{1}{2}\overline{{\cal{A}}}_{n}\overline{{\cal{D}}}_{m}\psi^{mn}-\frac{1}{2}\phi\psi-\phi^{m}\partial_{m}\psi

which is on the form that shows that this will correspond to the gauge fixing condition

𝒟¯mψmnsubscript¯𝒟𝑚superscript𝜓𝑚𝑛\displaystyle\overline{{\cal{D}}}_{m}\psi^{mn} =\displaystyle= 00\displaystyle 0

The BRST variations, which are inherited from the 5d MSYM supersymmetry upon twisting, read

δψmn𝛿subscript𝜓𝑚𝑛\displaystyle\delta\psi_{mn} =\displaystyle= mnsubscript𝑚𝑛\displaystyle{\cal{F}}_{mn}
δψm𝛿subscript𝜓𝑚\displaystyle\delta\psi_{m} =\displaystyle= 00\displaystyle 0
δψ𝛿𝜓\displaystyle\delta\psi =\displaystyle= ϕitalic-ϕ\displaystyle-\phi
δ𝒜¯m𝛿subscript¯𝒜𝑚\displaystyle\delta\overline{\cal{A}}_{m} =\displaystyle= 2iψm2𝑖subscript𝜓𝑚\displaystyle-2i\psi_{m}
δ𝒜m𝛿subscript𝒜𝑚\displaystyle\delta{\cal{A}}_{m} =\displaystyle= 00\displaystyle 0
δϕ𝛿italic-ϕ\displaystyle\delta\phi =\displaystyle= 00\displaystyle 0

While these fix the two-form gauge symmetry for ψmnsubscript𝜓𝑚𝑛\psi_{mn}, they still leave a residual gauge symmetry for Amsubscript𝐴𝑚A_{m} which we need to further BRST gauge fix. We do that in the usual fashion by adding the anticommuting c𝑐c and c¯¯𝑐\bar{c} ghosts and the auxiliary field B𝐵B for which we have the standard Yang-Mills BRST variations

δAmsuperscript𝛿subscript𝐴𝑚\displaystyle\delta^{\prime}A_{m} =\displaystyle= mcsubscript𝑚𝑐\displaystyle\partial_{m}c
δBsuperscript𝛿𝐵\displaystyle\delta^{\prime}B =\displaystyle= 00\displaystyle 0
δcsuperscript𝛿𝑐\displaystyle\delta^{\prime}c =\displaystyle= 00\displaystyle 0
δc¯superscript𝛿¯𝑐\displaystyle\delta^{\prime}\bar{c} =\displaystyle= iB𝑖𝐵\displaystyle iB

These are nilpotent, δ2=0superscriptsuperscript𝛿20{\delta^{\prime}}^{2}=0 and they commute with the supersymmetry variations, {δ,δ}=0𝛿superscript𝛿0\{\delta,\delta^{\prime}\}=0. A convenient choice for this second gauge fixing fermion is

Vsuperscript𝑉\displaystyle V^{\prime} =\displaystyle= ic¯(m𝒜m+ϕα2B)𝑖¯𝑐superscript𝑚subscript𝒜𝑚italic-ϕ𝛼2𝐵\displaystyle-i\bar{c}\left(\nabla^{m}{\cal{A}}_{m}+\phi-\frac{\alpha}{2}B\right)

We then get

δVsuperscript𝛿superscript𝑉\displaystyle\delta^{\prime}V^{\prime} =\displaystyle= B(m𝒜m+ϕ)α2B2+ic¯mmc𝐵superscript𝑚subscript𝒜𝑚italic-ϕ𝛼2superscript𝐵2𝑖¯𝑐superscript𝑚subscript𝑚𝑐\displaystyle B\left(\nabla^{m}{\cal{A}}_{m}+\phi\right)-\frac{\alpha}{2}B^{2}+i\bar{c}\nabla^{m}\partial_{m}c
δV𝛿𝑉\displaystyle\delta V =\displaystyle= 14¯mnmn+12ϕ2+ϕmmϕ14superscript¯𝑚𝑛subscript𝑚𝑛12superscriptitalic-ϕ2superscriptitalic-ϕ𝑚subscript𝑚italic-ϕ\displaystyle\frac{1}{4}\overline{\cal{F}}^{mn}{\cal{F}}_{mn}+\frac{1}{2}\phi^{2}+\phi^{m}\partial_{m}\phi
+i2ψmn(mψnnψm)+iψmmψ𝑖2superscript𝜓𝑚𝑛subscript𝑚subscript𝜓𝑛subscript𝑛subscript𝜓𝑚𝑖superscript𝜓𝑚subscript𝑚𝜓\displaystyle+\frac{i}{2}\psi^{mn}\left(\partial_{m}\psi_{n}-\partial_{n}\psi_{m}\right)+i\psi^{m}\partial_{m}\psi

and δV=δV=0𝛿superscript𝑉superscript𝛿𝑉0\delta V^{\prime}=\delta^{\prime}V=0. The full Lagrangian is

\displaystyle{\cal{L}} =\displaystyle= i8ϵmnpqrψmnpψqr+(δ+δ)(V+V)𝑖8superscriptitalic-ϵ𝑚𝑛𝑝𝑞𝑟subscript𝜓𝑚𝑛subscript𝑝subscript𝜓𝑞𝑟𝛿superscript𝛿𝑉superscript𝑉\displaystyle-\frac{i}{8}\epsilon^{mnpqr}\psi_{mn}\partial_{p}\psi_{qr}+\left(\delta+\delta^{\prime}\right)\left(V+V^{\prime}\right)

We now integrate out ϕitalic-ϕ\phi and then B𝐵B. We then end up with

\displaystyle{\cal{L}} =\displaystyle= 14Fmn2+12(α+1)(mAm)214superscriptsubscript𝐹𝑚𝑛212𝛼1superscriptsuperscript𝑚subscript𝐴𝑚2\displaystyle\frac{1}{4}F_{mn}^{2}+\frac{1}{2(\alpha+1)}\left(\nabla^{m}A_{m}\right)^{2}
14ϕmn212(mϕm)2+ic¯mmc14superscriptsubscriptitalic-ϕ𝑚𝑛212superscriptsuperscript𝑚subscriptitalic-ϕ𝑚2𝑖¯𝑐superscript𝑚subscript𝑚𝑐\displaystyle-\frac{1}{4}\phi_{mn}^{2}-\frac{1}{2}\left(\nabla^{m}\phi_{m}\right)^{2}+i\bar{c}\nabla^{m}\partial_{m}c
+i2ψmn(mψnnψm)+iψmmψ𝑖2superscript𝜓𝑚𝑛subscript𝑚subscript𝜓𝑛subscript𝑛subscript𝜓𝑚𝑖superscript𝜓𝑚subscript𝑚𝜓\displaystyle+\frac{i}{2}\psi^{mn}\left(\partial_{m}\psi_{n}-\partial_{n}\psi_{m}\right)+i\psi^{m}\partial_{m}\psi
i8ϵmnpqrψmnpψqr𝑖8superscriptitalic-ϵ𝑚𝑛𝑝𝑞𝑟subscript𝜓𝑚𝑛subscript𝑝subscript𝜓𝑞𝑟\displaystyle-\frac{i}{8}\epsilon^{mnpqr}\psi_{mn}\partial_{p}\psi_{qr}

We then take α=0𝛼0\alpha=0 to get the bosonic part of the action as

SBsubscript𝑆𝐵\displaystyle S_{B} =\displaystyle= 12(A,1A)12(ϕ,1ϕ)+i(c¯,0c)12𝐴subscript1𝐴12italic-ϕsubscript1italic-ϕ𝑖¯𝑐subscript0𝑐\displaystyle\frac{1}{2}(A,\triangle_{1}A)-\frac{1}{2}(\phi,\triangle_{1}\phi)+i(\bar{c},\triangle_{0}c)

The action for the fermions can be written in the form

SFsubscript𝑆𝐹\displaystyle S_{F} =\displaystyle= 12(Ψ,LΨ)12Ψ𝐿Ψ\displaystyle\frac{1}{2}(\Psi,L\Psi)

where Ψ:=(ψ2,ψ1,ψ0)TassignΨsuperscriptsubscript𝜓2subscript𝜓1subscript𝜓0𝑇\Psi:=(\psi_{2},\psi_{1},\psi_{0})^{T} and

L𝐿\displaystyle L =\displaystyle= i(dd0d0d0d0)\displaystyle i\left(\begin{array}[]{ccc}-*d&d&0\\ -d^{{\dagger}}&0&d\\ 0&-d^{{\dagger}}&0\end{array}\right)

is a hermitian operator that squares to

L2superscript𝐿2\displaystyle L^{2} =\displaystyle= (200010000)subscript2000subscript1000subscript0\displaystyle\left(\begin{array}[]{ccc}\triangle_{2}&0&0\\ 0&\triangle_{1}&0\\ 0&0&\triangle_{0}\end{array}\right)

We get the following contributions to the partition function,

Zϕsubscript𝑍italic-ϕ\displaystyle Z_{\phi} =\displaystyle= 1det1121superscriptsubscript112\displaystyle\frac{1}{\det{}^{\frac{1}{2}}\triangle_{1}}
ZYMsubscript𝑍𝑌𝑀\displaystyle Z_{YM} =\displaystyle= det0det112subscript0superscriptsubscript112\displaystyle\frac{\det{}\triangle_{0}}{\det{}^{\frac{1}{2}}\triangle_{1}}
ZFsubscript𝑍𝐹\displaystyle Z_{F} =\displaystyle= det214det114det014superscriptsubscript214superscriptsubscript114superscriptsubscript014\displaystyle\det{}^{\frac{1}{4}}\triangle_{2}\det{}^{\frac{1}{4}}\triangle_{1}\det{}^{\frac{1}{4}}\triangle_{0}

coming from the fields ϕmsubscriptitalic-ϕ𝑚\phi_{m}, the Yang-Mills gauge field Amsubscript𝐴𝑚A_{m}, and the fermionic fields ψ,ψm,ψmn𝜓subscript𝜓𝑚subscript𝜓𝑚𝑛\psi,\psi_{m},\psi_{mn} respectively. In these determinants, we can find zero modes. We will assume that b0=b5=1subscript𝑏0subscript𝑏51b_{0}=b_{5}=1 and that all the other Betti numbers are vanishing. This includes the lens spaces. In this case we have zero modes only coming from det0subscript0\det\triangle_{0}, which appears in both the fermionic part as well as the ghost part coming from the gauge fixing of the YM gauge potential. These are all fermionic ghost zero modes, which appear as a result of trying to gauge fix a gauge symmetry by a gauge fixing function that has a zero mode. These ghost zero modes come from fermionic ghosts associated with the gauge fixing of the fermionic two-form gauge symmetry and the YM gauge symmetry respectively. We should remove all ghost zero modes by further BRST gauge fixing. We present in detail how this is done in the appendix D. We define

Zosc=ZϕZYMZF=τoscsubscript𝑍𝑜𝑠𝑐subscript𝑍italic-ϕsubscriptsuperscript𝑍𝑌𝑀subscriptsuperscript𝑍𝐹subscript𝜏𝑜𝑠𝑐\displaystyle Z_{osc}=Z_{\phi}Z^{\prime}_{YM}Z^{\prime}_{F}=\sqrt{\tau_{osc}}

where τoscsubscript𝜏𝑜𝑠𝑐\tau_{osc} is defined in (4.5) below and primes are used to indicate that zero modes are taken out from the determinants. The full partition function is obtained by dividing by the volume of the gauge group bundle 𝒢𝒢{\cal{G}}

Z=1Vol(𝒢)Zosc𝑍1Vol𝒢subscript𝑍𝑜𝑠𝑐\displaystyle Z=\frac{1}{{\mbox{Vol}}({\cal{G}})}Z_{osc} (4.3)

In the appendix C we argue that when we take out a fermionic ghost zero mode that is associated with a gauge fixing that leaves a residual gauge symmetry, we shall divide by a corresponding volume factor of the unbroken gauge group. See also [8], [9], [10], [11] and the lectures [25]. Now this result has been applied to nonabelian 3d CS perturbation theory where we expand around some flat background gauge field and the unbroken gauge group refers to the stability group for that background gauge field. But here we have an abelian gauge group and perhaps then the background field may be thought of as the gauge field is zero, around which we ‘expand’ to quadratic order (that is, abelian theory with no interactions). That background gauge field being zero, does not break the abelian gauge group, so we have to divide by it. That is, we divide by the volume Vol(𝒢)Vol𝒢{\mbox{Vol}}({\cal{G}}). Now the Zoscsubscript𝑍𝑜𝑠𝑐Z_{osc} factor in the partition function has only a contribution coming from the oscillator modes since the zero modes have been taken out (or gauge fixed away). But we anticipate the full partition function will involve the Ray-Singer torsion which has both a zero mode part and an oscillator mode part. Indeed the zero mode contribution to the Ray-Singer torsion comes from the division by Vol(𝒢)Vol𝒢{\mbox{Vol}}({\cal{G}}). As explained in [25] (Eq. (3.64) in the arxiv version v5), we have

Vol(𝒢)Vol𝒢\displaystyle{\mbox{Vol}}({\cal{G}}) =\displaystyle= Vol(U(1))Vol(M5)12Vol𝑈1Volsuperscriptsubscript𝑀512\displaystyle{\mbox{Vol}}(U(1)){\mbox{Vol}}(M_{5})^{\frac{1}{2}}

so we get

Z=1Vol(U(1))τ𝑍1Vol𝑈1𝜏\displaystyle Z=\frac{1}{{\mbox{Vol}}(U(1))}\sqrt{\tau}

where τ=τzeroτosc𝜏subscript𝜏𝑧𝑒𝑟𝑜subscript𝜏𝑜𝑠𝑐\tau=\tau_{zero}\tau_{osc} is the Ray-Singer torsion, composed of the zero mode and the oscillator mode contributions

τzerosubscript𝜏𝑧𝑒𝑟𝑜\displaystyle\tau_{zero} =\displaystyle= 1Vol(M5)1Volsubscript𝑀5\displaystyle\frac{1}{{\mbox{Vol}}(M_{5})} (4.4)
τoscsubscript𝜏𝑜𝑠𝑐\displaystyle\tau_{osc} =\displaystyle= det212det052det132superscriptsubscript212superscriptsubscript052superscriptsubscript132\displaystyle\frac{\det{}^{\prime\frac{1}{2}}\triangle_{2}\det{}^{\prime\frac{5}{2}}\triangle_{0}}{\det{}^{\prime\frac{3}{2}}\triangle_{1}} (4.5)

The Ray-Singer torsion is independent of the volume, but the determinants do depend on the volume. If R𝑅R denotes a typical length scale of M5subscript𝑀5M_{5}, then all the Laplacians will scale like R2similar-tosuperscript𝑅2\triangle\sim R^{-2}, and so

detR2ζ(0)=R2bpsimilar-tosuperscript𝑅2subscript𝜁0superscript𝑅2subscript𝑏𝑝\displaystyle\det\triangle\sim R^{-2\zeta_{\triangle}(0)}=R^{2b_{p}}

where bp=dimHpsubscript𝑏𝑝dimensionsubscript𝐻𝑝b_{p}=\dim H_{p}. Both steps in the above relation are nontrivial. We refer to appendix B for more details. Let us now consider the oscillator mode contribution to the Ray-Singer torsion,

τosc=p=05(detp)(1)pp2p=05R(1)ppbpsubscript𝜏𝑜𝑠𝑐superscriptsubscriptproduct𝑝05superscriptsubscript𝑝superscript1𝑝𝑝2similar-tosuperscriptsubscriptproduct𝑝05superscript𝑅superscript1𝑝𝑝subscript𝑏𝑝\displaystyle\tau_{osc}=\prod_{p=0}^{5}\left(\det\triangle_{p}\right)^{-(-1)^{p}\frac{p}{2}}\sim\prod_{p=0}^{5}R^{-(-1)^{p}pb_{p}}

Thus this will have a nontrivial dependence on R𝑅R, that we need to cancel by multiplying by a zero mode contribution. Let us assume that b0=b5=1subscript𝑏0subscript𝑏51b_{0}=b_{5}=1 and all other Betti numbers are zero. Then

τoscR5similar-tosubscript𝜏𝑜𝑠𝑐superscript𝑅5\displaystyle\tau_{osc}\sim R^{5}

We thus need the zero mode contribution to be

τzero1Vol(M5)similar-tosubscript𝜏𝑧𝑒𝑟𝑜1Volsubscript𝑀5\displaystyle\tau_{zero}\sim\frac{1}{{\mbox{Vol}}(M_{5})}

to cancel the dependence on R𝑅R, as Vol(M5)R5similar-toVolsubscript𝑀5superscript𝑅5{\mbox{Vol}}(M_{5})\sim R^{5}. We verify explicitly this dependence on R𝑅R for the case that M5=L(p;1,1)subscript𝑀5𝐿𝑝11M_{5}=L(p;1,1) in the appendix A.1 where we get

τzerosubscript𝜏𝑧𝑒𝑟𝑜\displaystyle\tau_{zero} =\displaystyle= pπ3R5𝑝superscript𝜋3superscript𝑅5\displaystyle\frac{p}{\pi^{3}R^{5}}
τoscsubscript𝜏𝑜𝑠𝑐\displaystyle\tau_{osc} =\displaystyle= π3R5p3superscript𝜋3superscript𝑅5superscript𝑝3\displaystyle\frac{\pi^{3}R^{5}}{p^{3}}

The full partition function on L(p;1,1)𝐿𝑝11L(p;1,1) for abelian gauge group is now

Z=pVol(U(1))τ=1Vol(U(1))𝑍𝑝Vol𝑈1𝜏1Vol𝑈1\displaystyle Z=\frac{p}{{\mbox{Vol}}(U(1))}\sqrt{\tau}=\frac{1}{{\mbox{Vol}}(U(1))} (4.6)

Here the factor of p𝑝p comes from summing over all holonomy sectors =0,,p10𝑝1{{\ell}}=0,...,p-1. For abelian gauge group, the Weyl group is trivial and so we do not cut the sum at =(p1)/2absent𝑝12\;=(p-1)/2 as we do for SU(2)𝑆𝑈2SU(2) gauge group. Since all fields are in adjoint which is trivial for U(1)𝑈1U(1) gauge group, all holonomy sectors give rise to the same result and we just sum them up which leads to a factor of p𝑝p. The partition function is independent of R𝑅R and it is a topological invariant.

4.1 Nonabelian gauge group

For a nonabelian gauge group with all the fields transforming in the adjoint representation we introduce the covariant derivatives

𝒟msubscript𝒟𝑚\displaystyle{\cal{D}}_{m} =\displaystyle= mi𝒜msubscript𝑚𝑖subscript𝒜𝑚\displaystyle\nabla_{m}-i{\cal{A}}_{m}
𝒟¯msubscript¯𝒟𝑚\displaystyle\overline{\cal{D}}_{m} =\displaystyle= mi𝒜¯msubscript𝑚𝑖subscript¯𝒜𝑚\displaystyle\nabla_{m}-i\overline{\cal{A}}_{m}

The nonabelian fermionic Chern-Simons Lagrangian is given by

5d=FCS+δVsubscript5𝑑subscript𝐹𝐶𝑆𝛿𝑉\displaystyle{\cal{L}}_{5d}={\cal{L}}_{FCS}+\delta V

where

FCSsubscript𝐹𝐶𝑆\displaystyle{\cal{L}}_{FCS} =\displaystyle= i2ϵmnpqrTr(ψmn𝒟pψqr)𝑖2superscriptitalic-ϵ𝑚𝑛𝑝𝑞𝑟Trsubscript𝜓𝑚𝑛subscript𝒟𝑝subscript𝜓𝑞𝑟\displaystyle-\frac{i}{2}\epsilon^{mnpqr}{\mbox{Tr}}\left(\psi_{mn}{\cal{D}}_{p}\psi_{qr}\right)

and the gauge fixing fermion is chosen as

V𝑉\displaystyle V =\displaystyle= tr(12¯mnψmn12ϕψϕm𝒟mψ)tr12subscript¯𝑚𝑛superscript𝜓𝑚𝑛12italic-ϕ𝜓superscriptitalic-ϕ𝑚subscript𝒟𝑚𝜓\displaystyle{\mbox{tr}}\left(\frac{1}{2}\overline{{\cal{F}}}_{mn}\psi^{mn}-\frac{1}{2}\phi\psi-\phi^{m}{\cal{D}}_{m}\psi\right)

The 2-form BRST variations read

δψmn𝛿subscript𝜓𝑚𝑛\displaystyle\delta\psi_{mn} =\displaystyle= 12mn12subscript𝑚𝑛\displaystyle\frac{1}{2}{\cal{F}}_{mn}
δψm𝛿subscript𝜓𝑚\displaystyle\delta\psi_{m} =\displaystyle= 00\displaystyle 0
δψ𝛿𝜓\displaystyle\delta\psi =\displaystyle= ϕitalic-ϕ\displaystyle-\phi
δ𝒜¯m𝛿subscript¯𝒜𝑚\displaystyle\delta\overline{\cal{A}}_{m} =\displaystyle= 2iψm2𝑖subscript𝜓𝑚\displaystyle-2i\psi_{m}
δ𝒜m𝛿subscript𝒜𝑚\displaystyle\delta{\cal{A}}_{m} =\displaystyle= 00\displaystyle 0
δϕ𝛿italic-ϕ\displaystyle\delta\phi =\displaystyle= 00\displaystyle 0

As before these fix the two-form gauge symmetry for ψmnsubscript𝜓𝑚𝑛\psi_{mn}, and we need to fix the residual gauge symmetry for the one-form Amsubscript𝐴𝑚A_{m}. The standard Yang-Mills BRST variations would not commute with the above 2-form BRST variations, so instead we take these one-form BRST variations as

δ𝒜¯msuperscript𝛿subscript¯𝒜𝑚\displaystyle\delta^{\prime}\overline{\cal{A}}_{m} =\displaystyle= 00\displaystyle 0
δ𝒜msuperscript𝛿subscript𝒜𝑚\displaystyle\delta^{\prime}{\cal{A}}_{m} =\displaystyle= 𝒟mcsubscript𝒟𝑚𝑐\displaystyle{\cal{D}}_{m}c
δBsuperscript𝛿𝐵\displaystyle\delta^{\prime}B =\displaystyle= 00\displaystyle 0
δcsuperscript𝛿𝑐\displaystyle\delta^{\prime}c =\displaystyle= i2{c,c}𝑖2𝑐𝑐\displaystyle\frac{i}{2}\{c,c\}
δc¯superscript𝛿¯𝑐\displaystyle\delta^{\prime}\bar{c} =\displaystyle= iB𝑖𝐵\displaystyle iB

These are still nilpotent, δ2=0superscriptsuperscript𝛿20{\delta^{\prime}}^{2}=0 but now these also commute with the 2-form BRST variations, {δ,δ}=0𝛿superscript𝛿0\{\delta,\delta^{\prime}\}=0. A convenient choice for the second gauge fixing fermion is

Vsuperscript𝑉\displaystyle V^{\prime} =\displaystyle= tr(ic¯(m𝒜m+ϕα2B))tr𝑖¯𝑐superscript𝑚subscript𝒜𝑚italic-ϕ𝛼2𝐵\displaystyle{\mbox{tr}}\left(-i\bar{c}\left(\nabla^{m}{\cal{A}}_{m}+\phi-\frac{\alpha}{2}B\right)\right)

which gives

δVsuperscript𝛿superscript𝑉\displaystyle\delta^{\prime}V^{\prime} =\displaystyle= tr(B(mAmmϕm+ϕ)α2B2+ic¯m𝒟mc)tr𝐵superscript𝑚subscript𝐴𝑚superscript𝑚subscriptitalic-ϕ𝑚italic-ϕ𝛼2superscript𝐵2𝑖¯𝑐superscript𝑚subscript𝒟𝑚𝑐\displaystyle{\mbox{tr}}\left(B\left(\nabla^{m}A_{m}-\nabla^{m}\phi_{m}+\phi\right)-\frac{\alpha}{2}B^{2}+i\bar{c}\nabla^{m}{\cal{D}}_{m}c\right)
δVsuperscript𝛿𝑉\displaystyle\delta^{\prime}V =\displaystyle= 00\displaystyle 0
δV𝛿superscript𝑉\displaystyle\delta V^{\prime} =\displaystyle= 00\displaystyle 0
δV𝛿𝑉\displaystyle\delta V =\displaystyle= tr(14¯mnmn+12ϕ2ϕ𝒟mϕm)tr14superscript¯𝑚𝑛subscript𝑚𝑛12superscriptitalic-ϕ2italic-ϕsuperscript𝒟𝑚subscriptitalic-ϕ𝑚\displaystyle{\mbox{tr}}\left(\frac{1}{4}\overline{\cal{F}}^{mn}{\cal{F}}_{mn}+\frac{1}{2}\phi^{2}-\phi{\cal{D}}^{m}\phi_{m}\right)

Then integrating out ϕitalic-ϕ\phi puts ϕ=𝒟mϕmBitalic-ϕsuperscript𝒟𝑚subscriptitalic-ϕ𝑚𝐵\phi={\cal{D}}^{m}\phi_{m}-B and then integrating out B𝐵B puts B=11+α𝒟m𝒜m𝐵11𝛼superscript𝒟𝑚subscript𝒜𝑚B=\frac{1}{1+\alpha}{\cal{D}}^{m}{\cal{A}}_{m}. Thus after integrating out all the auxiliary fields, we get

\displaystyle{\cal{L}} =\displaystyle= tr(14¯mnmn12(𝒟mϕm)2+12(α+1)(𝒟m𝒜m)2ic¯m𝒟mc)tr14superscript¯𝑚𝑛subscript𝑚𝑛12superscriptsuperscript𝒟𝑚subscriptitalic-ϕ𝑚212𝛼1superscriptsuperscript𝒟𝑚subscript𝒜𝑚2𝑖¯𝑐superscript𝑚subscript𝒟𝑚𝑐\displaystyle{\mbox{tr}}\left(\frac{1}{4}\bar{\cal{F}}^{mn}{\cal{F}}_{mn}-\frac{1}{2}\left({\cal{D}}^{m}\phi_{m}\right)^{2}+\frac{1}{2(\alpha+1)}\left({\cal{D}}^{m}{\cal{A}}_{m}\right)^{2}-i\bar{c}\nabla^{m}{\cal{D}}_{m}c\right)
i8ϵmnpqrtr(ψmn𝒟pψqr)𝑖8superscriptitalic-ϵ𝑚𝑛𝑝𝑞𝑟trsubscript𝜓𝑚𝑛subscript𝒟𝑝subscript𝜓𝑞𝑟\displaystyle-\frac{i}{8}\epsilon^{mnpqr}{\mbox{tr}}\left(\psi_{mn}{\cal{D}}_{p}\psi_{qr}\right)

Preserving just one scalar real supercharge (or BRST charge), we can put the 5d FCS theory a generic five-manifold M5subscript𝑀5M_{5}, which has no isometries. Then supersymmetric field configurations satisfy

\displaystyle{\cal{F}} =\displaystyle= 00\displaystyle 0
dϕsuperscript𝑑italic-ϕ\displaystyle d^{{\dagger}}\phi =\displaystyle= 00\displaystyle 0

Here555Here we Wick rotate ϕmsubscriptitalic-ϕ𝑚\phi_{m} into iϕm𝑖subscriptitalic-ϕ𝑚i\phi_{m}, which, as we explained in [14], corresponds to Wick rotating time in the 6d theory. As we also explained there, BPS equations are better analyzed in this Euclidean theory.

\displaystyle{\cal{F}} =\displaystyle= Fidϕ+iϕ2𝐹𝑖𝑑italic-ϕ𝑖superscriptitalic-ϕ2\displaystyle F-id\phi+i\phi^{2}

so that the BPS equations become

F+iϕ2𝐹𝑖superscriptitalic-ϕ2\displaystyle F+i\phi^{2} =\displaystyle= 00\displaystyle 0
dϕ𝑑italic-ϕ\displaystyle d\phi =\displaystyle= 00\displaystyle 0
dϕsuperscript𝑑italic-ϕ\displaystyle d^{{\dagger}}\phi =\displaystyle= 00\displaystyle 0

This says that ϕitalic-ϕ\phi shall be a harmonic one-form. If we assume that the first cohomology group H1(M5,)superscript𝐻1subscript𝑀5H^{1}(M_{5},\mathbb{Z}) is trivial, this implies that

F𝐹\displaystyle F =\displaystyle= 00\displaystyle 0
ϕitalic-ϕ\displaystyle\phi =\displaystyle= 00\displaystyle 0

and the only BPS configurations are flat connections.

Let us now assume that M5subscript𝑀5M_{5} is a K-contact manifold. Then it has a unit normalizable Killing vector vmsuperscript𝑣𝑚v^{m}. if Gmnsubscript𝐺𝑚𝑛G_{mn} is the metric, then we define a contact one-form as κm=Gmnvnsubscript𝜅𝑚subscript𝐺𝑚𝑛superscript𝑣𝑛\kappa_{m}=G_{mn}v^{n} and unit norm means κmvm=1subscript𝜅𝑚superscript𝑣𝑚1\kappa_{m}v^{m}=1. Now in addition to the flat connections, we now also have contact instantons [26] as saddle points, satisfying

F𝐹\displaystyle F =\displaystyle= (κF)absent𝜅𝐹\displaystyle*(\kappa\wedge F)
ιvFsubscript𝜄𝑣𝐹\displaystyle\iota_{v}F =\displaystyle= 00\displaystyle 0

We can add further BRST exact terms which will enhance the supersymmetry to two supercharges [14] and then the contact instantons can become supersymmetric solutions. Yet these instanton configurations are never localization points since the BRST exact part of the action is always nonzero on these instantons. These contributions will become exponentially suppressed and in the localization limit their contribution to the partition function becomes zero. This may sound counter-intuitive since the Yang-Mills action evaluated on contact instantons is proportional to RT𝑅𝑇\frac{R}{T} where R𝑅R is the radius of S5/psuperscript𝑆5subscript𝑝S^{5}/\mathbb{Z}_{p} and T𝑇T is the radius of the time-circle along which we reduce from 6d to 5d. Small instantons correspond to Kaluza-Klein modes under dimensional reduction from 6d. This suggests that the value of the classical Yang-Mills action in 5d theory could have a physical interpretation as a Kaluza-Klein momentum. But this is in contradiction with the fact that the Yang-Mills term sits in the BRST exact part of the Lagrangian and the fact that we can rescale the BRST exact part at our wish without affecting any physical observables. Then the ratio RT𝑅𝑇\frac{R}{T} can not have any invariant significance. This is not a contradiction though, since nothing depends on radius R𝑅R since the theory is topological over S5/psuperscript𝑆5subscript𝑝S^{5}/\mathbb{Z}_{p} and the ratio RT𝑅𝑇\frac{R}{T} has no invariant physical meaning.

Assuming gauge group SU(2)𝑆𝑈2SU(2), then on S5/p=L(p;q1,q2)superscript𝑆5subscript𝑝𝐿𝑝subscript𝑞1subscript𝑞2S^{5}/\mathbb{Z}_{p}=L(p;q_{1},q_{2}) we can have holonomies

U𝑈\displaystyle U =\displaystyle= (e2πip00e2πip)matrixsuperscript𝑒2𝜋𝑖𝑝00superscript𝑒2𝜋𝑖𝑝\displaystyle\begin{pmatrix}e^{\frac{2\pi i{{\ell}}}{p}}&0\\ 0&e^{-\frac{2\pi i{{\ell}}}{p}}\end{pmatrix}

labeled by integers subject to the psubscript𝑝\mathbb{Z}_{p} identification +psimilar-to𝑝{{\ell}}\sim{{\ell}}+p. Unlike the case with 3d CS classical action when evaluated on a flat gauge field, which gives a phase when exponentiated, here we have the 5d YM action which is vanishing on a flat gauge field. The whole action is vanishing on the flat gauge field background.

Now we can apply the localization method as follows. We write the full Lagrangian as FCS+cδVtotsubscript𝐹𝐶𝑆𝑐𝛿subscript𝑉𝑡𝑜𝑡{\cal{L}}_{FCS}+c\delta V_{tot} where Vtot=V+Vsubscript𝑉𝑡𝑜𝑡𝑉superscript𝑉V_{tot}=V+V^{\prime} and c>0𝑐0c>0 is a parameter on which nothing depends. We may then take c𝑐c\rightarrow\infty. Then the path integral localizes to localization points where δVtot=0𝛿subscript𝑉𝑡𝑜𝑡0\delta V_{tot}=0 and its first derivative is zero. For this argument to work, we need δVtot0𝛿subscript𝑉𝑡𝑜𝑡0\delta V_{tot}\geq 0, which, we have assured ourselves, is the case. All the sectors with nontrivial holonomies are kept since at these localization points we have δVtot=0𝛿subscript𝑉𝑡𝑜𝑡0\delta V_{tot}=0.

It remains to compute the one-loop determinant for the fluctuations around flat gauge field backgrounds. We rescale all the fluctuation fields around the classical flat gauge field background by the factor 1/c1𝑐1/\sqrt{c}. This has the effect of rescaling the FCS term by 1/c1𝑐1/c. Taking c𝑐c large, we can neglect all higher order interaction terms in the fluctuation fields and only consider the one-loop approximation which becomes exact as we take c𝑐c to infinity. If we denote by dA=di[A,]subscript𝑑𝐴𝑑𝑖𝐴d_{A}=d-i[A,\cdot], then the fermionic operator L𝐿L becomes

L𝐿\displaystyle L =\displaystyle= i(dA/cdA0dA0dA0dA0)\displaystyle i\left(\begin{array}[]{ccc}-*d_{A}/c&d_{A}&0\\ -d_{A}^{{\dagger}}&0&d_{A}\\ 0&-d_{A}^{{\dagger}}&0\end{array}\right)

whose square is

L2superscript𝐿2\displaystyle L^{2} =\displaystyle= (dAdA/c2+dAdA00010000)superscriptsubscript𝑑𝐴subscript𝑑𝐴superscript𝑐2subscript𝑑𝐴superscriptsubscript𝑑𝐴000subscript1000subscript0\displaystyle\left(\begin{array}[]{ccc}d_{A}^{{\dagger}}d_{A}/c^{2}+d_{A}d_{A}^{{\dagger}}&0&0\\ 0&\triangle_{1}&0\\ 0&0&\triangle_{0}\end{array}\right)

if being understood that these Laplacians are given in terms of the background gauge field by dAdA+dAdAsubscript𝑑𝐴superscriptsubscript𝑑𝐴superscriptsubscript𝑑𝐴subscript𝑑𝐴d_{A}d_{A}^{{\dagger}}+d_{A}^{{\dagger}}d_{A}. Let us put C=1/c2𝐶1superscript𝑐2C=1/c^{2} that we would like take towards zero in order to localize the path integral. We now have to consider the following determinant

det(C2coex+2ex)=det(C2coex+1coex)=det(C2coex)det(1coex)𝐶superscriptsubscript2𝑐𝑜𝑒𝑥superscriptsubscript2𝑒𝑥𝐶superscriptsubscript2𝑐𝑜𝑒𝑥superscriptsubscript1𝑐𝑜𝑒𝑥𝐶superscriptsubscript2𝑐𝑜𝑒𝑥superscriptsubscript1𝑐𝑜𝑒𝑥\displaystyle\det\left(C\triangle_{2}^{coex}+\triangle_{2}^{ex}\right)=\det\left(C\triangle_{2}^{coex}+\triangle_{1}^{coex}\right)=\det\left(C\triangle_{2}^{coex}\right)\det\left(\triangle_{1}^{coex}\right)

We have the following result (see Eq. (B.2) in the appendix B.1)

det(Cp)=Cbpdetp𝐶subscript𝑝superscript𝐶subscript𝑏𝑝subscript𝑝\displaystyle\det\left(C\triangle_{p}\right)=C^{-b_{p}}\det\triangle_{p}

Also, from

det(Cp)𝐶subscript𝑝\displaystyle\det(C\triangle_{p}) =\displaystyle= det(Cpcoex)det(Cp1coex)𝐶superscriptsubscript𝑝𝑐𝑜𝑒𝑥𝐶superscriptsubscript𝑝1𝑐𝑜𝑒𝑥\displaystyle\det(C\triangle_{p}^{coex})\det(C\triangle_{p-1}^{coex})

we can iteratively deduce the scaling behavior of det(Cpcoex)𝐶superscriptsubscript𝑝𝑐𝑜𝑒𝑥\det\left(C\triangle_{p}^{coex}\right), starting with

det(C0coex)𝐶subscriptsuperscript𝑐𝑜𝑒𝑥0\displaystyle\det\left(C\triangle^{coex}_{0}\right) =\displaystyle= Cb0det0coexsuperscript𝐶subscript𝑏0subscriptsuperscript𝑐𝑜𝑒𝑥0\displaystyle C^{-b_{0}}\det\triangle^{coex}_{0}

which is valid simply because 0=0coexsubscript0superscriptsubscript0𝑐𝑜𝑒𝑥\triangle_{0}=\triangle_{0}^{coex}. Then we get iteratively

det(C1coex)𝐶subscriptsuperscript𝑐𝑜𝑒𝑥1\displaystyle\det\left(C\triangle^{coex}_{1}\right) =\displaystyle= Cb1+b0det1coexsuperscript𝐶subscript𝑏1subscript𝑏0subscriptsuperscript𝑐𝑜𝑒𝑥1\displaystyle C^{-b_{1}+b_{0}}\det\triangle^{coex}_{1}
det(C2coex)𝐶subscriptsuperscript𝑐𝑜𝑒𝑥2\displaystyle\det\left(C\triangle^{coex}_{2}\right) =\displaystyle= Cb2+b1b0det2coexsuperscript𝐶subscript𝑏2subscript𝑏1subscript𝑏0subscriptsuperscript𝑐𝑜𝑒𝑥2\displaystyle C^{-b_{2}+b_{1}-b_{0}}\det\triangle^{coex}_{2}

and so in particular we get

det(C2coex+2ex)=Cb2+b1b0det2𝐶superscriptsubscript2𝑐𝑜𝑒𝑥superscriptsubscript2𝑒𝑥superscript𝐶subscript𝑏2subscript𝑏1subscript𝑏0subscript2\displaystyle\det\left(C\triangle_{2}^{coex}+\triangle_{2}^{ex}\right)=C^{-b_{2}+b_{1}-b_{0}}\det\triangle_{2}

We then get the following contributions to the partition function,

Zϕsubscript𝑍italic-ϕ\displaystyle Z_{\phi} =\displaystyle= 1det1121superscriptsubscript112\displaystyle\frac{1}{\det{}^{\frac{1}{2}}\triangle_{1}}
ZYMsubscript𝑍𝑌𝑀\displaystyle Z_{YM} =\displaystyle= det0det112subscript0superscriptsubscript112\displaystyle\frac{\det{}\triangle_{0}}{\det{}^{\frac{1}{2}}\triangle_{1}}
ZFsubscript𝑍𝐹\displaystyle Z_{F} =\displaystyle= C(b2+b1b0)/4det214det114det014superscript𝐶subscript𝑏2subscript𝑏1subscript𝑏04superscriptsubscript214superscriptsubscript114superscriptsubscript014\displaystyle C^{(-b_{2}+b_{1}-b_{0})/4}\det{}^{\frac{1}{4}}\triangle_{2}\det{}^{\frac{1}{4}}\triangle_{1}\det{}^{\frac{1}{4}}\triangle_{0}

These are the contributions from the fields ϕmsubscriptitalic-ϕ𝑚\phi_{m}, the Yang-Mills part Amsubscript𝐴𝑚A_{m}, and the fermionic part, respectively. We find zero modes, which we take out. Then multiplying the contributions together gives the following oscillator mode contribution to the partition function

Zosc=ZϕZYMZF=C(b2+b1b0)/4(τosc)12subscript𝑍𝑜𝑠𝑐subscript𝑍italic-ϕsubscriptsuperscript𝑍𝑌𝑀subscriptsuperscript𝑍𝐹superscript𝐶subscript𝑏2subscript𝑏1subscript𝑏04superscriptsubscript𝜏𝑜𝑠𝑐12\displaystyle Z_{osc}=Z_{\phi}Z^{\prime}_{YM}Z^{\prime}_{F}=C^{(-b_{2}+b_{1}-b_{0})/4}(\tau_{osc})^{\frac{1}{2}}

It now seems that ZoscCb0/4=cb0/2similar-tosubscript𝑍𝑜𝑠𝑐superscript𝐶subscript𝑏04superscript𝑐subscript𝑏02Z_{osc}\sim C^{-b_{0}/4}=c^{b_{0}/2} depends on a coefficient c𝑐c that we will take to infinity. (We will assume that b1=b2=0subscript𝑏1subscript𝑏20b_{1}=b_{2}=0, and let us also recall that b0=dimHAsubscript𝑏0dimensionsubscript𝐻𝐴b_{0}=\dim H_{A} where A𝐴A represents the background gauge field and HAsubscript𝐻𝐴H_{A} is the isotropy group preserving this background gauge field.) No matter how small we make C𝐶C, the contribution from the FCS-term can not be neglected when we compute the one-loop determinant. The normal situation in supersymmetric localization is that we can neglect the contribution from the original action when we compute the one-loop determinants and only the BRST exact terms contribute to the one-loop determinants. The quadratic terms that sit in the original action are suppressed by the factor C=1/c2𝐶1superscript𝑐2C=1/c^{2} compared to the quadratic terms that sit in the BRST exact terms. Here the situation is different since there are no terms that are quadratic in ψ2subscript𝜓2\psi_{2} in the BRST exact part of our Lagrangian. The leading quadratic term in ψ2subscript𝜓2\psi_{2} is in the original FCS-term, so this contribution can not be neglected when we compute the one-loop determinant, and hence our dependence on the coefficient c𝑐c. But in our computation we forgot to take into account a corresponding rescaling in the path integral measure. If we rescale all the modes, then this has no effect on the measure. Since our manifold is compact, the modes form a countable set. Let us label the modes by an integer n𝑛n. Then the measure receives a product factor n1c=1subscriptproduct𝑛1𝑐1\prod_{n\in\mathbb{Z}}\frac{1}{\sqrt{c}}=1 by rescaling all the modes by a factor of 1c1𝑐\frac{1}{\sqrt{c}}. When we take out a zero mode n=0𝑛0n=0, the product starts to depend on c𝑐c as n01c=csubscriptproduct𝑛01𝑐𝑐\prod_{n\neq 0}\frac{1}{\sqrt{c}}=\sqrt{c}. Since there are b0subscript𝑏0b_{0} zero modes in total, we get from the path integral measure the factor cb0/2superscript𝑐subscript𝑏02c^{b_{0}/2}. The precise way to see this is by expanding in mode functions that are eigenmodes of the laplace operator 0subscript0\triangle_{0} on M5subscript𝑀5M_{5}, and then use zeta function regularization for the infinite product over nonvanishing eigenmodes (nonzero modes) of 0subscript0\triangle_{0},

n01c=(1c)ζ0(0)=cb0/2subscriptproduct𝑛01𝑐superscript1𝑐subscript𝜁subscript00superscript𝑐subscript𝑏02\displaystyle\prod_{n\neq 0}\frac{1}{\sqrt{c}}=\left(\frac{1}{\sqrt{c}}\right)^{\zeta_{\triangle_{0}}(0)}=c^{b_{0}/2}

where in the last step we used the Minakshisundaram-Pleijel theorem ζ0(0)=b0subscript𝜁subscript00subscript𝑏0\zeta_{\triangle_{0}}(0)=-b_{0} (see appendix B.1). Now the exponent has the wrong sign – we need cb0/2superscript𝑐subscript𝑏02c^{-b_{0}/2} to cancel the above c𝑐c-dependence from the determinants. But this is actually what we have since the contribution to the zero modes comes from fermionic fields rather than from bosonic fields. For fermionic fields we have the following property d(1cψ)=cdψ𝑑1𝑐𝜓𝑐𝑑𝜓d(\frac{1}{\sqrt{c}}\psi)=\sqrt{c}d\psi when we rescale the field, the differential in the measure is rescaled by the inverse factor. So for fermionic fields we will encounter the product n0c=1csubscriptproduct𝑛0𝑐1𝑐\prod_{n\neq 0}\sqrt{c}=\frac{1}{\sqrt{c}} instead and this gives us the desired factor of cb0/2superscript𝑐subscript𝑏02c^{-b_{0}/2} that cancels out all the dependence on c𝑐c.

Our Laplacian operators are defined in a flat gauge field background. Alternatively we consider fields that are satisfying twisted boundary conditions along the Hopf circle. We shall sum over all gauge inequivalent flat gauge field backgrounds. For SU(2)𝑆𝑈2SU(2) gauge group and lens space L(p;q1,q2)𝐿𝑝subscript𝑞1subscript𝑞2L(p;q_{1},q_{2}) that amounts to a sum over flat gauge fields that we constructed explicitly in section 2 and which are labeled by =0,1,,(p1)/201𝑝12{{\ell}}=0,1,...,(p-1)/2 assuming p𝑝p is odd. We need to multiple by a factor Zzerosubscript𝑍𝑧𝑒𝑟𝑜Z_{zero} that is obtained by dividing by the volume factor [25]

Vol(A)Volsubscriptsuperscript𝐴\displaystyle{\mbox{Vol}}({\cal{H}}_{A^{{{\ell}}}}) =\displaystyle= Vol(HA)(Vol(M5))12dim(HA)Volsubscript𝐻superscript𝐴superscriptVolsubscript𝑀512dimensionsubscript𝐻superscript𝐴\displaystyle{\mbox{Vol}}(H_{A^{{{\ell}}}})({\mbox{Vol}}(M_{5}))^{\frac{1}{2}\dim(H_{A^{{{\ell}}}})}

where HAsubscript𝐻superscript𝐴H_{A^{{{\ell}}}} denotes the subgroup of the gauge group that leaves the background gauge field Asuperscript𝐴A^{{{\ell}}} invariant. For gauge group SU(2)𝑆𝑈2SU(2), these volume factors are

Vol(𝒢)Vol𝒢\displaystyle{\mbox{Vol}}({\cal{G}}) =\displaystyle= Vol(SU(2))(Vol(L(p;q1,q2)))32Vol𝑆𝑈2superscriptVol𝐿𝑝subscript𝑞1subscript𝑞232\displaystyle{\mbox{Vol}}(SU(2))({\mbox{Vol}}(L(p;q_{1},q_{2})))^{\frac{3}{2}}

for =00{{\ell}}=0 and

Vol()Vol\displaystyle{\mbox{Vol}}({\cal{H}}) =\displaystyle= Vol(U(1))(Vol(L(p;q1,q2)))12Vol𝑈1superscriptVol𝐿𝑝subscript𝑞1subscript𝑞212\displaystyle{\mbox{Vol}}(U(1))({\mbox{Vol}}(L(p;q_{1},q_{2})))^{\frac{1}{2}}

for >00{{\ell}}>0. The Ray-Singer torsion for SU(2)𝑆𝑈2SU(2) gauge group is given by (1.8), which we can separate into a zero mode factor and an oscillator mode factor for each {{\ell}} as

τ0,SU(2)subscript𝜏0𝑆𝑈2\displaystyle\tau_{0,SU(2)} =\displaystyle= [τ0,zeroτ0,osc]3superscriptdelimited-[]subscript𝜏0𝑧𝑒𝑟𝑜subscript𝜏0𝑜𝑠𝑐3\displaystyle\left[\tau_{0,zero}\tau_{0,osc}\right]^{3}
τ,SU(2)subscript𝜏𝑆𝑈2\displaystyle\tau_{{{\ell}},SU(2)} =\displaystyle= [τ0,zeroτ0,osc]τ2τ2delimited-[]subscript𝜏0𝑧𝑒𝑟𝑜subscript𝜏0𝑜𝑠𝑐subscript𝜏2subscript𝜏2\displaystyle\left[\tau_{0,zero}\tau_{0,osc}\right]\tau_{2{{\ell}}}\tau_{-2{{\ell}}}

where τ0,zero=1Vol(L(p;q1,q2))subscript𝜏0𝑧𝑒𝑟𝑜1Vol𝐿𝑝subscript𝑞1subscript𝑞2\tau_{0,zero}=\frac{1}{{\mbox{Vol}}(L(p;q_{1},q_{2}))}. The resulting partition function becomes precisely of the form presented in Eq (1.9). Unlike the case for 3d CS, for 5d FCS this one-loop result is exact by the localization principle.

It remains the question of the normalization of the 5d FCS term, or equivalently, the question of the radius of the SU(2)𝑆𝑈2SU(2) gauge group. We will not fully solve this problem, but will be able to derive the dependence on kFsubscript𝑘𝐹k_{F}, where we normalize the 5d FCS term as

ikF2tr(ψ2dAψ2)=ikF2(ψ2,dAψ2)\displaystyle\frac{ik_{F}}{2}\int{\mbox{tr}}(\psi_{2}\wedge d_{A}\psi_{2})=\frac{ik_{F}}{2}(\psi_{2},*d_{A}\psi_{2})

This leads to a determinant

det(ikFdA)1/2superscript𝑖subscript𝑘𝐹subscript𝑑𝐴12\displaystyle\det(ik_{F}*d_{A})^{1/2} (4.9)

in the space of non-harmonic two-forms. The kFsubscript𝑘𝐹k_{F}-dependence of the determinant (4.9) can be inferred from

dAdAabsentsubscript𝑑𝐴subscript𝑑𝐴\displaystyle*d_{A}*d_{A} =\displaystyle= 2coexsuperscriptsubscript2𝑐𝑜𝑒𝑥\displaystyle\triangle_{2}^{coex}

together with the formula

det(kF22coex)1/4superscriptsuperscriptsubscript𝑘𝐹2superscriptsubscript2𝑐𝑜𝑒𝑥14\displaystyle\det(k_{F}^{2}\triangle_{2}^{coex})^{1/4} =\displaystyle= kF(b2+b1b0)/2det(2coex)1/4superscriptsubscript𝑘𝐹subscript𝑏2subscript𝑏1subscript𝑏02superscriptsuperscriptsubscript2𝑐𝑜𝑒𝑥14\displaystyle k_{F}^{(-b_{2}+b_{1}-b_{0})/2}\det(\triangle_{2}^{coex})^{1/4}

Then for b2=b1=0subscript𝑏2subscript𝑏10b_{2}=b_{1}=0 and b0subscript𝑏0b_{0} being the dimension of the unbroken gauge group, this leads to the kFsubscript𝑘𝐹k_{F}-dependence kFb0/2similar-toabsentsuperscriptsubscript𝑘𝐹subscript𝑏02\sim k_{F}^{-b_{0}/2}. Unlike the dependence on c𝑐c above, the dependence on kFsubscript𝑘𝐹k_{F} is genuine. The reason why, is that kFsubscript𝑘𝐹k_{F} multiplies the FCS term, while c𝑐c multiplies the BRST exact terms. Also here we did not need to rescale the field and so we did not change the path integral measure as we did when we extracted the dependence on c𝑐c.

Let us compare this dependence on kFsubscript𝑘𝐹k_{F} with the dependence on k𝑘k for 3d CS perturbation theory. There we have the CS term

k4π(B,dAB)\displaystyle\frac{k}{4\pi}(B,*d_{A}B)

where B𝐵B represents a fluctuation of the gauge field around the background A𝐴A. This leads to the determinant

1det(kdA)1𝑘subscript𝑑𝐴\displaystyle\frac{1}{\det(k*d_{A})} =\displaystyle= 1k(b1+b0)/2det(dA)\displaystyle\frac{1}{k^{(-b_{1}+b_{0})/2}\det(*d_{A})}

and we get the k𝑘k-dependence kb0/2similar-toabsentsuperscript𝑘subscript𝑏02\sim k^{-b_{0}/2} (assuming that b1=0subscript𝑏10b_{1}=0), which is of the same form as we found for 5d FCS. That suggests that as an alternative to our localization compuation, we should also be able to study 5d FCS theory using perturbation theory in a small coupling 1/kF1subscript𝑘𝐹1/k_{F}.

5 Higher-dimensional knot theory

In [27], abelian 3d CS was used to describe links by associating a Wilson loop with a link or a knot. Here we would like to generalize this to links and knots that are made of closed 2d surfaces embedded in a 5d manifold. The first question we should ask is what would be the definition of the Wilson surface that we should associate with such a 2d surface?

At least if the gauge group is abelian, we can dualize the 5d YM gauge potential A1subscript𝐴1A_{1} into a bosonic two-form B2subscript𝐵2B_{2} by taking dB2=dA1dB_{2}=*dA_{1}. We can then use this two-form to define a Wilson surface as

W(Σ1)𝑊subscriptΣ1\displaystyle W(\Sigma_{1}) =\displaystyle= exp(ieiΣiB)𝑖subscript𝑒𝑖subscriptsubscriptΣ𝑖𝐵\displaystyle\exp\left(ie_{i}\int_{\Sigma_{i}}B\right)

where eisubscript𝑒𝑖e_{i} is the charge associated with the loop whose trajectory forms the surface ΣisubscriptΣ𝑖\Sigma_{i}. However, this can not be used to describe knots made of closed 2d surfaces (surfaces that form topologically nontrivial links and knot configurations) in 5d. The reason is that the linking number is anti-symmetric under exchange of two surfaces in 5d.

Instead we should use a fermionic two-form ψ2subscript𝜓2\psi_{2}. We associate an anticommuting parameter θisubscript𝜃𝑖\theta_{i} with each surface ΣisubscriptΣ𝑖\Sigma_{i} embedded in some five-manifold. One may think on θisubscript𝜃𝑖\theta_{i} as the fermionic analog of electric charge. The product θiψ2subscript𝜃𝑖subscript𝜓2\theta_{i}\psi_{2} should have scaling dimension zero, since only then can we form a fermionic Wilson surface as

W(Σi)𝑊subscriptΣ𝑖\displaystyle W(\Sigma_{i}) =\displaystyle= exp(θiΣiψ2)subscript𝜃𝑖subscriptsubscriptΣ𝑖subscript𝜓2\displaystyle\exp\left(\theta_{i}\int_{\Sigma_{i}}\psi_{2}\right)

This Wilson surface becomes a unitary operator without the insertion of any extra factor of i𝑖i in the exponent, if θisubscript𝜃𝑖\theta_{i} and ψ2subscript𝜓2\psi_{2} are real and anticommuting. In 5superscript5\mathbb{R}^{5} we can have two 2d closed oriented surfaces linking each other. The Gauss linking formula gives the linking number of two such surfaces ΣisubscriptΣ𝑖\Sigma_{i} and ΣjsubscriptΣ𝑗\Sigma_{j} as

k(Σi,Σj)ksubscriptΣ𝑖subscriptΣ𝑗\displaystyle{\ell{\mbox{k}}}(\Sigma_{i},\Sigma_{j}) similar-to\displaystyle\sim Σi𝑑xmdxnΣj𝑑ypdyqϵmnpqrxryr|xy|5subscriptsubscriptΣ𝑖differential-dsuperscript𝑥𝑚𝑑superscript𝑥𝑛subscriptsubscriptΣ𝑗differential-dsuperscript𝑦𝑝𝑑superscript𝑦𝑞subscriptitalic-ϵ𝑚𝑛𝑝𝑞𝑟superscript𝑥𝑟superscript𝑦𝑟superscript𝑥𝑦5\displaystyle\int_{\Sigma_{i}}dx^{m}\wedge dx^{n}\int_{\Sigma_{j}}dy^{p}\wedge dy^{q}\epsilon_{mnpqr}\frac{x^{r}-y^{r}}{|x-y|^{5}}

This linking number is anti-symmetric in 5d

k(Σi,Σj)ksubscriptΣ𝑖subscriptΣ𝑗\displaystyle{\ell{\mbox{k}}}(\Sigma_{i},\Sigma_{j}) =\displaystyle= k(Σj,Σi)ksubscriptΣ𝑗subscriptΣ𝑖\displaystyle-{\ell{\mbox{k}}}(\Sigma_{j},\Sigma_{i})

We now notice that fermionic Chern-Simons leads to the propagator

ψmn(x)ψpq(y)delimited-⟨⟩subscript𝜓𝑚𝑛𝑥subscript𝜓𝑝𝑞𝑦\displaystyle\left<\psi_{mn}(x)\psi_{pq}(y)\right> similar-to\displaystyle\sim ϵmnpqrxryr|xy|5subscriptitalic-ϵ𝑚𝑛𝑝𝑞𝑟superscript𝑥𝑟superscript𝑦𝑟superscript𝑥𝑦5\displaystyle\epsilon_{mnpqr}\frac{x^{r}-y^{r}}{|x-y|^{5}}

and so we can compute this linking number from the expectation value of two Wilson surfaces

k(Σi,Σj)ksubscriptΣ𝑖subscriptΣ𝑗\displaystyle{\ell{\mbox{k}}}(\Sigma_{i},\Sigma_{j}) =\displaystyle= W(Σi)W(Σj)delimited-⟨⟩𝑊subscriptΣ𝑖𝑊subscriptΣ𝑗\displaystyle\left<W(\Sigma_{i})W(\Sigma_{j})\right>

More generally, if we have a disconnected set of surfaces ΣisubscriptΣ𝑖\Sigma_{i}, we define our Wilson surface as

W(Σ1,,Σn)=exp(i=1nθiΣiψ2)=exp(i=1nθiψ2δΣi)=exp(ψ2δΣf)𝑊subscriptΣ1subscriptΣ𝑛superscriptsubscript𝑖1𝑛superscript𝜃𝑖subscriptsubscriptΣ𝑖subscript𝜓2superscriptsubscript𝑖1𝑛subscript𝜃𝑖subscript𝜓2subscript𝛿subscriptΣ𝑖subscript𝜓2subscriptsuperscript𝛿𝑓Σ\displaystyle W(\Sigma_{1},...,\Sigma_{n})=\exp\left(\sum_{i=1}^{n}\theta^{i}\int_{\Sigma_{i}}\psi_{2}\right)=\exp\left(\sum_{i=1}^{n}\theta_{i}\int\psi_{2}\wedge\delta_{\Sigma_{i}}\right)=\exp\left(\int\psi_{2}\wedge\delta^{f}_{\Sigma}\right)

Here we define a fermionic Poincare dual as

δΣfsubscriptsuperscript𝛿𝑓Σ\displaystyle\delta^{f}_{\Sigma} :=assign\displaystyle:= i=1nθiδΣisuperscriptsubscript𝑖1𝑛subscript𝜃𝑖subscript𝛿subscriptΣ𝑖\displaystyle\sum_{i=1}^{n}\theta_{i}\delta_{\Sigma_{i}}

and δΣisubscript𝛿subscriptΣ𝑖\delta_{\Sigma_{i}} are the usual Poincare duals of ΣisubscriptΣ𝑖\Sigma_{i}, defined by

ψ2δΣisubscript𝜓2subscript𝛿subscriptΣ𝑖\displaystyle\int\psi_{2}\wedge\delta_{\Sigma_{i}} =\displaystyle= Σiψ2subscriptsubscriptΣ𝑖subscript𝜓2\displaystyle\int_{\Sigma_{i}}\psi_{2}

We can now compute the expectation value of this generalized Wilson surface,

W(Σ1,,Σn)=𝒟ψ2ei2(ψ2dψ22iψ2δΣf)delimited-⟨⟩𝑊subscriptΣ1subscriptΣ𝑛𝒟subscript𝜓2superscript𝑒𝑖2subscript𝜓2𝑑subscript𝜓22𝑖subscript𝜓2subscriptsuperscript𝛿𝑓Σ\displaystyle\left<W(\Sigma_{1},...,\Sigma_{n})\right>=\int{\cal{D}}\psi_{2}e^{\frac{i}{2}\int\left(\psi_{2}\wedge d\psi_{2}-2i\psi_{2}\wedge\delta^{f}_{\Sigma}\right)}

by shifting the fermionic field as

χ𝜒\displaystyle\chi =\displaystyle= ψiδΣ𝜓𝑖subscript𝛿Σ\displaystyle\psi-i\delta_{\Sigma}

We then complete the square, and we get

W(Σ1,,Σn)=1eδΣfδDf=1eθiθjδΣiδDj=1eθiθjk(Σi,Σj)delimited-⟨⟩𝑊subscriptΣ1subscriptΣ𝑛delimited-⟨⟩1superscript𝑒superscriptsubscript𝛿Σ𝑓superscriptsubscript𝛿𝐷𝑓delimited-⟨⟩1superscript𝑒subscript𝜃𝑖subscript𝜃𝑗subscript𝛿subscriptΣ𝑖subscript𝛿subscript𝐷𝑗delimited-⟨⟩1superscript𝑒subscript𝜃𝑖subscript𝜃𝑗ksubscriptΣ𝑖subscriptΣ𝑗\displaystyle\left<W(\Sigma_{1},...,\Sigma_{n})\right>=\left<1\right>e^{\int\delta_{\Sigma}^{f}\wedge\delta_{D}^{f}}=\left<1\right>e^{\theta_{i}\theta_{j}\int\delta_{\Sigma_{i}}\wedge\delta_{D_{j}}}=\left<1\right>e^{\theta_{i}\theta_{j}{{\ell}}{\rm k}(\Sigma_{i},\Sigma_{j})}

where we define the linking number as

k(Σi,Σj)ksubscriptΣ𝑖subscriptΣ𝑗\displaystyle{\ell{\mbox{k}}}(\Sigma_{i},\Sigma_{j}) =\displaystyle= δΣiδDjsubscript𝛿subscriptΣ𝑖subscript𝛿subscript𝐷𝑗\displaystyle\int\delta_{\Sigma_{i}}\wedge\delta_{D_{j}}

with Di=Misubscript𝐷𝑖subscript𝑀𝑖\partial D_{i}=M_{i}. Now we can understand the absence of framing dependent factor in the partition function. The self-intersection is simply removed by the anticommuting property θiθj=θjθisubscript𝜃𝑖subscript𝜃𝑗subscript𝜃𝑗subscript𝜃𝑖\theta_{i}\theta_{j}=-\theta_{j}\theta_{i}. So we do not need to consider the issue of framing to define the otherwise ambiguous self-intersection numbers.

6 Uplift to six dimensions

The maximal twist of the M5 brane theory amounts to identifying the R-symmetry group SO(5)𝑆𝑂5SO(5) with the SO(5)SO(1,5)𝑆𝑂5𝑆𝑂15SO(5)\subset SO(1,5) in the Lorentz group. Once this twist is done, we can preserve one scalar supercharge on any Lorentzian six-manifold M6=×M5subscript𝑀6subscript𝑀5M_{6}=\mathbb{R}\times M_{5} with metric

ds2𝑑superscript𝑠2\displaystyle ds^{2} =\displaystyle= dt2+Gmndxmdxn𝑑superscript𝑡2subscript𝐺𝑚𝑛𝑑superscript𝑥𝑚𝑑superscript𝑥𝑛\displaystyle-dt^{2}+G_{mn}dx^{m}dx^{n}

The action is

SBsubscript𝑆𝐵\displaystyle S_{B} =\displaystyle= 12(12(B2,~2B2)12(B0,~0B)+(c¯,~0c))1212subscript𝐵2subscript~2subscript𝐵212subscript𝐵0subscript~0𝐵¯𝑐subscript~0𝑐\displaystyle\frac{1}{2}\left(-\frac{1}{2}(B_{2},\widetilde{\triangle}_{2}B_{2})-\frac{1}{2}(B_{0},\widetilde{\triangle}_{0}B)+(\overline{c},\widetilde{\triangle}_{0}c)\right)
+i(b¯1,~1b1)𝑖subscript¯𝑏1subscript~1subscript𝑏1\displaystyle+i(\overline{b}_{1},\widetilde{\triangle}_{1}b_{1})
Sϕsubscript𝑆italic-ϕ\displaystyle S_{\phi} =\displaystyle= 12(ϕ˙,ϕ˙)12(ϕ,1ϕ)12˙italic-ϕ˙italic-ϕ12italic-ϕsubscript1italic-ϕ\displaystyle\frac{1}{2}(\dot{\phi},\dot{\phi})-\frac{1}{2}(\phi,\triangle_{1}\phi)
Sψsubscript𝑆𝜓\displaystyle S_{\psi} =\displaystyle= i2(ψ0,ψ˙0)i2(ψ1,ψ˙1)i2(ψ2,ψ˙2)𝑖2subscript𝜓0subscript˙𝜓0𝑖2subscript𝜓1subscript˙𝜓1𝑖2subscript𝜓2subscript˙𝜓2\displaystyle-\frac{i}{2}(\psi_{0},\dot{\psi}_{0})-\frac{i}{2}(\psi_{1},\dot{\psi}_{1})-\frac{i}{2}(\psi_{2},\dot{\psi}_{2})
+i(ψ1,dψ0)+i(ψ2,dψ1)𝑖subscript𝜓1𝑑subscript𝜓0𝑖subscript𝜓2𝑑subscript𝜓1\displaystyle+i(\psi_{1},d\psi_{0})+i(\psi_{2},d\psi_{1})
i2(ψ2,dψ2)\displaystyle-\frac{i}{2}(\psi_{2},*d\psi_{2})

where ~~\widetilde{\triangle} denotes a 6d Laplacian. Here b1subscript𝑏1b_{1} and b¯1subscript¯𝑏1\overline{b}_{1} are ghosts for the two-form gauge field B2subscript𝐵2B_{2}, and B𝐵B, c𝑐c and c¯¯𝑐\overline{c} are ghosts-for-ghosts [28, 29]. The relative coefficients of the full action are fixed by supersymmetry and by the requirement that, upon dimensional reduction from S1×5superscript𝑆1superscript5S^{1}\times\mathbb{R}^{5} down to 5superscript5\mathbb{R}^{5}, we get 5d SYM theory (on flat space, twisting is trivial) with canonically normalized fields. For this twisted theory we have just one real supercharge. The term for the two bosonic ghosts c𝑐c and c¯¯𝑐\overline{c} is not of the standard form. But by using the identity

𝑑x𝑑yeλxy=𝑑x𝑑yeiλ2(x2+y2)=2πiλdifferential-d𝑥differential-d𝑦superscript𝑒𝜆𝑥𝑦differential-d𝑥differential-d𝑦superscript𝑒𝑖𝜆2superscript𝑥2superscript𝑦22𝜋𝑖𝜆\displaystyle\int dxdye^{\lambda xy}=\int dxdye^{\frac{i\lambda}{2}(x^{2}+y^{2})}=\frac{2\pi i}{\lambda}

this term can be replaced as

(c¯,~0c)¯𝑐subscript~0𝑐\displaystyle(\overline{c},\widetilde{\triangle}_{0}c) \displaystyle\rightarrow i2(c,~0c)+i2(c¯,~0c¯)𝑖2𝑐subscript~0𝑐𝑖2¯𝑐subscript~0¯𝑐\displaystyle\frac{i}{2}(c,\widetilde{\triangle}_{0}c)+\frac{i}{2}(\overline{c},\widetilde{\triangle}_{0}\overline{c})

without changing the value of the path integral.

The action for the fermions can be written in the form

Sψsubscript𝑆𝜓\displaystyle S_{\psi} =\displaystyle= i2(Ψ,Ψ˙)+12(Ψ,LΨ)𝑖2Ψ˙Ψ12Ψ𝐿Ψ\displaystyle-\frac{i}{2}(\Psi,\dot{\Psi})+\frac{1}{2}(\Psi,L\Psi)

where Ψ:=(ψ2,ψ1,ψ0)TassignΨsuperscriptsubscript𝜓2subscript𝜓1subscript𝜓0𝑇\Psi:=(\psi_{2},\psi_{1},\psi_{0})^{T} and

L𝐿\displaystyle L =\displaystyle= i(dd0d0d0d0)\displaystyle i\left(\begin{array}[]{ccc}-*d&d&0\\ -d^{{\dagger}}&0&d\\ 0&-d^{{\dagger}}&0\end{array}\right)

is a hermitian operator that squares to

L2superscript𝐿2\displaystyle L^{2} =\displaystyle= (200010000)subscript2000subscript1000subscript0\displaystyle\left(\begin{array}[]{ccc}\triangle_{2}&0&0\\ 0&\triangle_{1}&0\\ 0&0&\triangle_{0}\end{array}\right)

We will now compute the Witten index using the path integral quantization. Since there are zero modes that we take out, we need to be careful with normalization of the path integral. Our normalization of the path integral will be that which for finite dimensional integrals corresponds to

dx2πeλ2x2𝑑𝑥2𝜋superscript𝑒𝜆2superscript𝑥2\displaystyle\int\frac{dx}{\sqrt{2\pi}}e^{-\frac{\lambda}{2}x^{2}} =\displaystyle= 1λ1𝜆\displaystyle\frac{1}{\sqrt{\lambda}}
𝑑ψ𝑑ψ¯eλψ¯ψdifferential-d𝜓differential-d¯𝜓superscript𝑒𝜆¯𝜓𝜓\displaystyle\int d\psi d\overline{\psi}e^{\lambda\overline{\psi}\psi} =\displaystyle= λ𝜆\displaystyle\lambda

In other words, we will get determinants without any extra multiplicative factors, when the action is canonically normalized. It turns out that all terms in our action have the canonical normalization. One can see this by computing the Dirac brackets. One then find the canonical commutation relations, which means the kinetic terms in our action are canonically normalized. This also amounts to no extra factors appear when we compute the path integral and get determinants. We get from the path integral the following results

Zψsubscript𝑍𝜓\displaystyle Z_{\psi} =\displaystyle= det(it+L)12superscript𝑖subscript𝑡𝐿12\displaystyle\det\left(i\partial_{t}+L\right)^{\frac{1}{2}}
=\displaystyle= det(t2+2)14det(t2+1)14det(t2+0)14superscriptsubscriptsuperscript2𝑡subscript214superscriptsubscriptsuperscript2𝑡subscript114superscriptsubscriptsuperscript2𝑡subscript014\displaystyle\det\left(\partial^{2}_{t}+\triangle_{2}\right)^{\frac{1}{4}}\det\left(\partial^{2}_{t}+\triangle_{1}\right)^{\frac{1}{4}}\det\left(\partial^{2}_{t}+\triangle_{0}\right)^{\frac{1}{4}}
Zϕsubscript𝑍italic-ϕ\displaystyle Z_{\phi} =\displaystyle= 1det(t2+1)121superscriptsuperscriptsubscript𝑡2subscript112\displaystyle\frac{1}{\det\left(\partial_{t}^{2}+\triangle_{1}\right)^{\frac{1}{2}}}
ZB+subscript𝑍superscript𝐵\displaystyle Z_{B^{+}} =\displaystyle= (det(~1)det(~2)12det(~0)12det(i~0))12superscriptsubscript~1superscriptsubscript~212superscriptsubscript~012𝑖subscript~012\displaystyle\left(\frac{\det\left(\widetilde{\triangle}_{1}\right)}{\det\left(\widetilde{\triangle}_{2}\right)^{\frac{1}{2}}\det\left(\widetilde{\triangle}_{0}\right)^{\frac{1}{2}}\det\left(-i\widetilde{\triangle}_{0}\right)}\right)^{\frac{1}{2}}
=\displaystyle= det(t2+1)14det(t2+2)14det(t2+0)14superscriptsubscriptsuperscript2𝑡subscript114superscriptsubscriptsuperscript2𝑡subscript214superscriptsubscriptsuperscript2𝑡subscript014\displaystyle\frac{\det\left(\partial^{2}_{t}+\triangle_{1}\right)^{\frac{1}{4}}}{\det\left(\partial^{2}_{t}+\triangle_{2}\right)^{\frac{1}{4}}\det\left(\partial^{2}_{t}+\triangle_{0}\right)^{\frac{1}{4}}}

In reality the M5 brane action has a fixed coupling constant, and the canonical normalization of the action is not correct. However, when b2(M6)=b3(M6)=0subscript𝑏2subscript𝑀6subscript𝑏3subscript𝑀60b_{2}(M_{6})=b_{3}(M_{6})=0, those effects caused by selfduality disappear and we may assume the action has been canonically normalized by an appropriate rescaling of the fields.

For the two-form B𝐵B there is a G=U(1)𝐺𝑈1G=U(1) gauge symmetry, and we have to factor out the volume Vol(𝒢)Vol𝒢{\mbox{Vol}}({\cal{G}}) of the corresponding G-bundle 𝒢𝒢{\cal{G}} over the six-manifold when we compute the path integral over all gauge redundant field configurations. But after BRST gauge fixing, such a volume is factored out from the path integral, and then canceled by dividing the path integral by Vol(𝒢)Vol𝒢{\mbox{Vol}}({\cal{G}}). The upshot is that we never see Vol(𝒢)Vol𝒢{\mbox{Vol}}({\cal{G}}) in the final result after the cancellation of these volumes has taken place. If we multiply together all contributions, we find that all determinants cancel and we are left with

I=ZB+ZϕZψ=1𝐼subscript𝑍superscript𝐵subscript𝑍italic-ϕsubscript𝑍𝜓1\displaystyle I=Z_{B^{+}}Z_{\phi}Z_{\psi}=1

We note that there are both fermionic as well as bosonic zero modes. They appear in the determinants det(t2+0)14superscriptsubscriptsuperscript2𝑡subscript014\det\left(\partial^{2}_{t}+\triangle_{0}\right)^{\frac{1}{4}} in the fermionic contribution as well as in the contribution coming from B+superscript𝐵B^{+}. These zero modes are canceled since these determinants exactly cancel. Hence we do not need to remove ghost zero modes by hand and consequently we do not divide by an extra volume factor Vol(𝒢)Vol𝒢{\mbox{Vol}}({\cal{G}}) as we did in the corresponding abelian 5d theory, Eq (4.3), where we took out ghost zero modes by hand.

Now we have an interesting mismatch between the 6d Witten index666It is the Witten index since we assume periodic boundary conditions for the fermions around the time circle in the path integral. I=1𝐼1I=1 and the corresponding 5d partition function given in eq (4.6). A conjecture is that 5d MSYM is precisely the same thing as 6d (2,0) theory on a circle [30], [31]. Here we have only one supercharge, so we are not addressing the conjecture in its original form which keeps 161616 supercharges. But our theories are nevertheless related, so it is interesting to find a mismatch here. The origin of this mismatch lies in how a selfdual 2-form in 6d reduces to a Yang-Mills gauge field in 5d, which we can demonstrate explicitly only by assuming the gauge group is abelian.

7 Dimensional reduction of selfdual two-form

The 6d and 5d partition functions of a 2-form and of a 1-form potential, respectively, are

Z6d,osc(2)superscriptsubscript𝑍6𝑑𝑜𝑠𝑐2\displaystyle Z_{6d,osc}^{(2)} =\displaystyle= det~1det~212det~032superscriptsubscript~1superscriptsubscript~212superscriptsubscript~032\displaystyle\frac{\det{}^{\prime}\widetilde{\triangle}_{1}}{\det{}^{\prime\frac{1}{2}}\widetilde{\triangle}_{2}\det{}^{\prime\frac{3}{2}}\widetilde{\triangle}_{0}}
Z5d,osc(1)superscriptsubscript𝑍5𝑑𝑜𝑠𝑐1\displaystyle Z_{5d,osc}^{(1)} =\displaystyle= det0det112superscriptsubscript0superscriptsubscript112\displaystyle\frac{\det{}^{\prime}\triangle_{0}}{\det{}^{\prime\frac{1}{2}}\triangle_{1}}

Gauge fixing shall be extended to include any ghost zero modes as well, which then gauge fixing will take out as indicated by the primes. If we dimensionally reduce a selfdual 2-form in 6d, down to 5d, we get a 1-form gauge potential in 5d. Let us expand the p𝑝p-form Laplace operator on the Euclidean six-manifold M6=S1×M5subscript𝑀6superscript𝑆1subscript𝑀5M_{6}=S^{1}\times M_{5} as777Our sign convention is such that =mmsubscript𝑚subscript𝑚\triangle=-\partial_{m}\partial_{m} on 5superscript5\mathbb{R}^{5}.

~psubscript~𝑝\displaystyle\widetilde{\triangle}_{p} =\displaystyle= τ2+psuperscriptsubscript𝜏2subscript𝑝\displaystyle-\partial_{\tau}^{2}+\triangle_{p}

We then note the relation

det~p(τ=0)superscriptsubscript~𝑝subscript𝜏0\displaystyle\det{}^{\prime}\widetilde{\triangle}_{p}(\partial_{\tau}=0) =\displaystyle= detpdetp1superscriptsubscript𝑝superscriptsubscript𝑝1\displaystyle\det{}^{\prime}\triangle_{p}\det{}^{\prime}\triangle_{p-1}

This means that if we put τ=0subscript𝜏0\partial_{\tau}=0 to get the dimensionally reduced theory, then the 6d partition function of the selfdual 2-form reduces to

Z6d,osc(2+)(τ=0)subscriptsuperscript𝑍limit-from26𝑑𝑜𝑠𝑐subscript𝜏0\displaystyle Z^{(2+)}_{6d,osc}(\partial_{\tau}=0) =\displaystyle= det114det214det014superscriptsubscript114superscriptsubscript214superscriptsubscript014\displaystyle\frac{\det{}^{\prime\frac{1}{4}}\triangle_{1}}{\det{}^{\prime\frac{1}{4}}\triangle_{2}\det{}^{\prime\frac{1}{4}}\triangle_{0}}

where now the Laplacians are on M5subscript𝑀5M_{5}, and where we took the square root of the non-chiral two-form partition function. Naively we would expect to get the partition function of the 5d Maxwell theory, Z5dsubscript𝑍5𝑑Z_{5d}. To see whether this is really true, let us form the ratio,

Z6d,osc(2+)(τ=0)Z5d,osc(1)subscriptsuperscript𝑍limit-from26𝑑𝑜𝑠𝑐subscript𝜏0subscriptsuperscript𝑍15𝑑𝑜𝑠𝑐\displaystyle\frac{Z^{(2+)}_{6d,osc}(\partial_{\tau}=0)}{Z^{(1)}_{5d,osc}} =\displaystyle= det134det214det054superscriptsubscript134superscriptsubscript214superscriptsubscript054\displaystyle\frac{\det{}^{\prime\frac{3}{4}}\triangle_{1}}{\det{}^{\prime\frac{1}{4}}\triangle_{2}\det{}^{\prime\frac{5}{4}}\triangle_{0}}

This quantity is related to the oscillator mode contribution to the Ray-Singer torsion of M5subscript𝑀5M_{5}, which is defined as

τosc(Md)subscript𝜏𝑜𝑠𝑐subscript𝑀𝑑\displaystyle\tau_{osc}(M_{d}) =\displaystyle= p=0d(detp)(1)pp2superscriptsubscriptproduct𝑝0𝑑superscriptsuperscriptsubscript𝑝superscript1𝑝𝑝2\displaystyle\prod_{p=0}^{d}(\det{}^{\prime}\triangle_{p})^{-(-1)^{p}\frac{p}{2}}

where d=5𝑑5d=5 is the dimension of the manifold Mdsubscript𝑀𝑑M_{d} in our case. By using the relation detp=det5psuperscriptsubscript𝑝superscriptsubscript5𝑝\det{}^{\prime}\triangle_{p}=\det{}^{\prime}\triangle_{5-p} which follows from the fact that * commutes with the Laplacian and maps a p𝑝p-form to a (5p)5𝑝(5-p)-form in a one-to-one fashion, the Ray-Singer torsion becomes

τosc(M5)subscript𝜏𝑜𝑠𝑐subscript𝑀5\displaystyle\tau_{osc}(M_{5}) =\displaystyle= det212det052det132superscriptsubscript212superscriptsubscript052superscriptsubscript132\displaystyle\frac{\det{}^{\prime\frac{1}{2}}\triangle_{2}\det{}^{\prime\frac{5}{2}}\triangle_{0}}{\det{}^{\prime\frac{3}{2}}\triangle_{1}}

We now see that

Z6d,osc(2+)(τ=0)subscriptsuperscript𝑍limit-from26𝑑𝑜𝑠𝑐subscript𝜏0\displaystyle Z^{(2+)}_{6d,osc}(\partial_{\tau}=0) =\displaystyle= 1τosc(M5)Z5d,osc(1)1subscript𝜏𝑜𝑠𝑐subscript𝑀5subscriptsuperscript𝑍15𝑑𝑜𝑠𝑐\displaystyle\frac{1}{\sqrt{\tau_{osc}(M_{5})}}Z^{(1)}_{5d,osc} (7.1)

This relation does not explain the emergence of the zero mode contribution to the Ray-Singer torsion. The relation that we wish to have reads

Z6d(2+)(τ=0)subscriptsuperscript𝑍limit-from26𝑑subscript𝜏0\displaystyle Z^{(2+)}_{6d}(\partial_{\tau}=0) =\displaystyle= 1τosc(M5)τzero(M5)Z5d(1)1subscript𝜏𝑜𝑠𝑐subscript𝑀5subscript𝜏𝑧𝑒𝑟𝑜subscript𝑀5subscriptsuperscript𝑍15𝑑\displaystyle\frac{1}{\sqrt{\tau_{osc}(M_{5})\tau_{zero}(M_{5})}}Z^{(1)}_{5d}

Then we use that τzero(M5)=1Vol(M5)subscript𝜏𝑧𝑒𝑟𝑜subscript𝑀51Volsubscript𝑀5\tau_{zero}(M_{5})=\frac{1}{{\mbox{Vol}}(M_{5})} and we get

Z6d(2+)(τ=0)subscriptsuperscript𝑍limit-from26𝑑subscript𝜏0\displaystyle Z^{(2+)}_{6d}(\partial_{\tau}=0) =\displaystyle= 1τosc(M5)Vol(M5)Z5d(1)1subscript𝜏𝑜𝑠𝑐subscript𝑀5Volsubscript𝑀5subscriptsuperscript𝑍15𝑑\displaystyle\frac{1}{\sqrt{\tau_{osc}(M_{5})}}\sqrt{{\mbox{Vol}}(M_{5})}Z^{(1)}_{5d}

Indeed, we have argued that the full supersymmetric partition functions are given by

Z5d(1)superscriptsubscript𝑍5𝑑1\displaystyle Z_{5d}^{(1)} =\displaystyle= 1Vol(M5)Z5,osc(1)1Volsubscript𝑀5superscriptsubscript𝑍5𝑜𝑠𝑐1\displaystyle\frac{1}{\sqrt{{\mbox{Vol}}(M_{5})}}Z_{5,osc}^{(1)}
Z6d(2+)superscriptsubscript𝑍6𝑑limit-from2\displaystyle Z_{6d}^{(2+)} =\displaystyle= Z6d,osc(2+)superscriptsubscript𝑍6𝑑𝑜𝑠𝑐limit-from2\displaystyle Z_{6d,osc}^{(2+)}

We divide the 5d partition function by the volume of the gauge group bundle since we take out gauge zero modes. For 6d case the zero modes are canceling out so we do not divide by a correponding volume there. Now this leads us back to the relation (7.1).

As we show in appendix F, this can be generalized to selfdual 2k2𝑘2k-form potential in 4k+24𝑘24k+2 dimensions for k=1,2,3,𝑘123k=1,2,3,... where we have

Z4k+2,osc(2k+)(τ=0)subscriptsuperscript𝑍limit-from2𝑘4𝑘2𝑜𝑠𝑐subscript𝜏0\displaystyle Z^{(2k+)}_{4k+2,osc}(\partial_{\tau}=0) =\displaystyle= 1τosc(M4k+1)Z4k+1,osc(2k1)1subscript𝜏𝑜𝑠𝑐subscript𝑀4𝑘1subscriptsuperscript𝑍2𝑘14𝑘1𝑜𝑠𝑐\displaystyle\frac{1}{\sqrt{\tau_{osc}(M_{4k+1})}}Z^{(2k-1)}_{4k+1,osc}

In 4k4𝑘4k dimensions (k=1,2,3,𝑘123k=1,2,3,...) things work differently. While in 4k+24𝑘24k+2 dimensions we have 2=1*^{2}=1 in Lorentzian signature, in 4k4𝑘4k dimensions we have 2=1*^{2}=1 in Euclidean signature. For a selfdual 2k12𝑘12k-1-form potential in 4k4𝑘4k dimensions the relation is

Z4k,osc(2k1,+)(τ=0)subscriptsuperscript𝑍2𝑘14𝑘𝑜𝑠𝑐subscript𝜏0\displaystyle Z^{(2k-1,+)}_{4k,osc}(\partial_{\tau}=0) =\displaystyle= τosc(M4k1)Z4k1,osc(2k2)subscript𝜏𝑜𝑠𝑐subscript𝑀4𝑘1subscriptsuperscript𝑍2𝑘24𝑘1𝑜𝑠𝑐\displaystyle\sqrt{\tau_{osc}(M_{4k-1})}Z^{(2k-2)}_{4k-1,osc}

Let us now assume we reduce a selfdual gauge field in 4d down to a scalar field in 3d. Now the abelian scalar in 3d is not a gauge field and does not require gauge fixing, but the gauge field in 4d does require gauge fixing. So now we shall divide the volume factor on the 4d side, not on the reduced 3d side. So we have the relations

Z4(1,+)subscriptsuperscript𝑍14\displaystyle Z^{(1,+)}_{4} =\displaystyle= 1Vol(M4)Z4,osc(1,+)1Volsubscript𝑀4subscriptsuperscript𝑍14𝑜𝑠𝑐\displaystyle\frac{1}{\sqrt{{\mbox{Vol}}(M_{4})}}Z^{(1,+)}_{4,osc}
Z3(0)subscriptsuperscript𝑍03\displaystyle Z^{(0)}_{3} =\displaystyle= Z3(0)subscriptsuperscript𝑍03\displaystyle Z^{(0)}_{3}

Since Vol(M4)=R×Vol(M3)Volsubscript𝑀4𝑅Volsubscript𝑀3{\mbox{Vol}}(M_{4})=R\times{\mbox{Vol}}(M_{3}), the volume factor now combines with τoscsubscript𝜏𝑜𝑠𝑐\tau_{osc} into the full Ray-Singer torsion up to a factor of R𝑅R, the radius of the circle along which we reduce. Note that it appears we should not not take the square root of the above volume factor as one might have expected when the gauge field is selfdual.

Finding the nonabelian generalization of this dimensional reduction will be very interesting. We believe that the 6d Witten index is I=1𝐼1I=1 for any nonabelian gauge group. Knowing the 5d partition function, we may infer that the mismatch comes from reducing a nonabelian selfdual two-form to 5d nonabelian YM gauge field. This can give some clues about what is the nonabelian selfdual two-form.

Acknowledgements

We would like to thank Dongmin Gang, Luca Grigoulo, Jeong-Hyuck Park, Domenico Seminara, and Masahito Yamazaki for enlightening discussions. D.B. was supported in part by IBS-R018-D2 and NRF Grant 2017R1A2B4003095. A.G. was supported by the grant “Geometry and Physics” from the Knut and Alice Wallenberg foundation and IBS-R018-D2.

Appendix A The Ray-Singer torsion

Lecture notes on the Ray-Singer torsion are [19, 20]. Here we summarize what we need for our purposes. Let M𝑀M be a compact smooth oriented manifold of dimension d𝑑d without boundaries. Let ωpisuperscriptsubscript𝜔𝑝𝑖\omega_{p}^{i} be a basis of harmonic p𝑝p-forms on M𝑀M, with corresponding dual cycles Cpisuperscriptsubscript𝐶𝑝𝑖C_{p}^{i} in the homology of M𝑀M. We define the inner product of two p𝑝p-forms as

(ω,η)𝜔𝜂\displaystyle(\omega,\eta) =\displaystyle= Mωη\displaystyle\int_{M}\omega\wedge*\eta

We now have a matrix (ωpi,ωpj)subscriptsuperscript𝜔𝑖𝑝subscriptsuperscript𝜔𝑗𝑝(\omega^{i}_{p},\omega^{j}_{p}) for each p𝑝p (which may be empty if there are no harmonic p𝑝p-forms on M𝑀M). We define the Ray-Singer torsion as

τ(M)𝜏𝑀\displaystyle\tau(M) =\displaystyle= τzero(M)τosc(M)subscript𝜏𝑧𝑒𝑟𝑜𝑀subscript𝜏𝑜𝑠𝑐𝑀\displaystyle\tau_{zero}(M)\tau_{osc}(M)

where

τosc(M)subscript𝜏𝑜𝑠𝑐𝑀\displaystyle\tau_{osc}(M) =\displaystyle= p=0d(detp)(1)pp2superscriptsubscriptproduct𝑝0𝑑superscriptsuperscriptsubscript𝑝superscript1𝑝𝑝2\displaystyle\prod_{p=0}^{d}(\det{}^{\prime}\triangle_{p})^{-(-1)^{p}\frac{p}{2}}
τzero(M)subscript𝜏𝑧𝑒𝑟𝑜𝑀\displaystyle\tau_{zero}(M) =\displaystyle= p=0d(vp)(1)p12superscriptsubscriptproduct𝑝0𝑑superscriptsubscript𝑣𝑝superscript1𝑝12\displaystyle\prod_{p=0}^{d}(v_{p})^{-(-1)^{p}\frac{1}{2}}

and where

vpsubscript𝑣𝑝\displaystyle v_{p} =\displaystyle= det(ωpi,ωpj)superscriptsubscript𝜔𝑝𝑖superscriptsubscript𝜔𝑝𝑗\displaystyle\det(\omega_{p}^{i},\omega_{p}^{j})

A second expression for the Ray-Singer torsion can be obtained if we decompose the Laplacian in the space of non-harmonic forms as

psubscript𝑝\displaystyle\triangle_{p} =\displaystyle= pcoex+pexsuperscriptsubscript𝑝𝑐𝑜𝑒𝑥superscriptsubscript𝑝𝑒𝑥\displaystyle\triangle_{p}^{coex}+\triangle_{p}^{ex}

Here

pcoexsuperscriptsubscript𝑝𝑐𝑜𝑒𝑥\displaystyle\triangle_{p}^{coex} =\displaystyle= dpdpsuperscriptsubscript𝑑𝑝subscript𝑑𝑝\displaystyle d_{p}^{{\dagger}}d_{p}
pexsuperscriptsubscript𝑝𝑒𝑥\displaystyle\triangle_{p}^{ex} =\displaystyle= dp1dp1subscript𝑑𝑝1superscriptsubscript𝑑𝑝1\displaystyle d_{p-1}d_{p-1}^{{\dagger}}

where dp=(dp):Ωp+1Ωp:superscriptsubscript𝑑𝑝superscriptsubscript𝑑𝑝subscriptΩ𝑝1subscriptΩ𝑝d_{p}^{{\dagger}}=\left(d_{p}\right)^{{\dagger}}:\Omega_{p+1}\rightarrow\Omega_{p}. We have

detpsubscript𝑝\displaystyle\det\triangle_{p} =\displaystyle= detpexdetpcoexsuperscriptsubscript𝑝𝑒𝑥superscriptsubscript𝑝𝑐𝑜𝑒𝑥\displaystyle\det\triangle_{p}^{ex}\det\triangle_{p}^{coex}

Since ωp1=dωp2+dηp+ωp1harmsubscript𝜔𝑝1𝑑subscript𝜔𝑝2superscript𝑑subscript𝜂𝑝superscriptsubscript𝜔𝑝1𝑎𝑟𝑚\omega_{p-1}=d\omega_{p-2}+d^{{\dagger}}\eta_{p}+\omega_{p-1}^{harm}, we have dωp1=ddηp𝑑subscript𝜔𝑝1𝑑superscript𝑑subscript𝜂𝑝d\omega_{p-1}=dd^{{\dagger}}\eta_{p} and hence only the coexact part of ωp1subscript𝜔𝑝1\omega_{p-1} contributes. Therefore

detpexsuperscriptsubscript𝑝𝑒𝑥\displaystyle\det\triangle_{p}^{ex} =\displaystyle= detp1coexsuperscriptsubscript𝑝1𝑐𝑜𝑒𝑥\displaystyle\det\triangle_{p-1}^{coex} (A.1)

and so we have

detpsubscript𝑝\displaystyle\det\triangle_{p} =\displaystyle= detpcoexdetp1coexsuperscriptsubscript𝑝𝑐𝑜𝑒𝑥superscriptsubscript𝑝1𝑐𝑜𝑒𝑥\displaystyle\det\triangle_{p}^{coex}\det\triangle_{p-1}^{coex} (A.2)

If we use (A.2) and also note that dcoexsuperscriptsubscript𝑑𝑐𝑜𝑒𝑥\triangle_{d}^{coex} is trivial, we get

τoscsubscript𝜏𝑜𝑠𝑐\displaystyle\tau_{osc} =\displaystyle= p=0d(detpcoex)12(1)psuperscriptsubscriptproduct𝑝0𝑑superscriptsuperscriptsubscript𝑝𝑐𝑜𝑒𝑥12superscript1𝑝\displaystyle\prod_{p=0}^{d}(\det\triangle_{p}^{coex})^{\frac{1}{2}(-1)^{p}}

If ωpsubscript𝜔𝑝\omega_{p} is coexact, then ωpabsentsubscript𝜔𝑝*\omega_{p} is exact and we have and isomorphism between coexact p𝑝p forms and exact dp𝑑𝑝d-p forms that implies that

detpcoexsuperscriptsubscript𝑝𝑐𝑜𝑒𝑥\displaystyle\det\triangle_{p}^{coex} =\displaystyle= detdpexsuperscriptsubscript𝑑𝑝𝑒𝑥\displaystyle\det\triangle_{d-p}^{ex}

Moreover, by using the relation (A.1) we find that

detpcoexsuperscriptsubscript𝑝𝑐𝑜𝑒𝑥\displaystyle\det\triangle_{p}^{coex} =\displaystyle= detdp1coexsuperscriptsubscript𝑑𝑝1𝑐𝑜𝑒𝑥\displaystyle\det\triangle_{d-p-1}^{coex} (A.3)

We can use this relation to get

p=0d(detpcoex)12(1)psuperscriptsubscriptproduct𝑝0𝑑superscriptsuperscriptsubscript𝑝𝑐𝑜𝑒𝑥12superscript1𝑝\displaystyle\prod_{p=0}^{d}(\det\triangle_{p}^{coex})^{\frac{1}{2}(-1)^{p}} =\displaystyle= q=0d(detqcoex)12(1)q(1)d1superscriptsubscriptproduct𝑞0𝑑superscriptsuperscriptsubscript𝑞𝑐𝑜𝑒𝑥12superscript1𝑞superscript1𝑑1\displaystyle\prod_{q=0}^{d}(\det\triangle_{q}^{coex})^{\frac{1}{2}(-1)^{q}(-1)^{d-1}}

Hence, for even dimensions d𝑑d we have τosc=1subscript𝜏𝑜𝑠𝑐1\tau_{osc}=1. Also if d𝑑d is even, we get τzero=1subscript𝜏𝑧𝑒𝑟𝑜1\tau_{zero}=1 by Poincare duality, and so τ=1𝜏1\tau=1.

A third expression for the Ray-Singer torsion is expressed in terms of the Minakshisundaram-Pleijel zeta function of the Laplacian acting on p𝑝p-forms on M𝑀M,

ζ(s)subscript𝜁𝑠\displaystyle\zeta_{\triangle}(s) =\displaystyle= λi0λissubscriptsubscript𝜆𝑖0superscriptsubscript𝜆𝑖𝑠\displaystyle\sum_{\lambda_{i}\neq 0}\lambda_{i}^{-s}

Here the sum runs over nonzero eigenvalues λisubscript𝜆𝑖\lambda_{i} of the Laplacian. Then we define

F(s)𝐹𝑠\displaystyle F(s) =\displaystyle= k=0d(1)kk2ζk(s)superscriptsubscript𝑘0𝑑superscript1𝑘𝑘2subscript𝜁subscript𝑘𝑠\displaystyle\sum_{k=0}^{d}-\frac{(-1)^{k}k}{2}\zeta_{\triangle_{k}}(s)

and we have

τoscsubscript𝜏𝑜𝑠𝑐\displaystyle\tau_{osc} =\displaystyle= eF(0)superscript𝑒superscript𝐹0\displaystyle e^{-F^{\prime}(0)}

We now review the proof for the metric-independence of the Ray-Singer torsion that can be found in [20]. The proof shows the necessary structure of τzerosubscript𝜏𝑧𝑒𝑟𝑜\tau_{zero} in the presence of zero modes. It shows that τzerosubscript𝜏𝑧𝑒𝑟𝑜\tau_{zero} has a certain ambiguity. This ambiguity can be fixed by imposing an extra condition, as we do in Eq. (A.4).

We assume that the dimension d𝑑d is odd and we begin by assuming that there are no zero modes. We note that

ζk(s)subscript𝜁subscript𝑘𝑠\displaystyle\zeta_{\triangle_{k}}(s) =\displaystyle= 1Γ(s)0𝑑tts1trΩk(etk)1Γ𝑠superscriptsubscript0differential-d𝑡superscript𝑡𝑠1subscripttrsubscriptΩ𝑘superscript𝑒𝑡subscript𝑘\displaystyle\frac{1}{\Gamma(s)}\int_{0}^{\infty}dtt^{s-1}{\mbox{tr}}_{\Omega_{k}}\left(e^{-t\triangle_{k}}\right)

and so we have

F(s)𝐹𝑠\displaystyle F(s) =\displaystyle= 1Γ(s)0𝑑tts1k=0d(1)kk2trΩk(etk)1Γ𝑠superscriptsubscript0differential-d𝑡superscript𝑡𝑠1superscriptsubscript𝑘0𝑑superscript1𝑘𝑘2subscripttrsubscriptΩ𝑘superscript𝑒𝑡subscript𝑘\displaystyle\frac{1}{\Gamma(s)}\int_{0}^{\infty}dtt^{s-1}\sum_{k=0}^{d}-\frac{(-1)^{k}k}{2}{\mbox{tr}}_{\Omega_{k}}\left(e^{-t\triangle_{k}}\right)

Let u𝑢u parametrize a one-parameter family of metrics. The Hodge duality operators depends on the metric and hence on u𝑢u. We may emphasize that by writing it as =usubscript𝑢*=*_{u}. We now define the operator

α=˙:=udduu\displaystyle\alpha=*\dot{*}:=*_{u}\frac{d}{du}*_{u}

We have

(1)k(dk)\displaystyle(-1)^{k(d-k)}** =\displaystyle= 11\displaystyle 1
dsuperscript𝑑\displaystyle d^{{\dagger}} =\displaystyle= (1)dk+d+1d\displaystyle(-1)^{dk+d+1}*d*

which for d𝑑d odd yields

\displaystyle** =\displaystyle= 11\displaystyle 1
dsuperscript𝑑\displaystyle d^{{\dagger}} =\displaystyle= (1)kd\displaystyle(-1)^{k}*d*

Consequently ˙=α\dot{*}*=-\alpha and

˙˙\displaystyle\dot{\triangle} =\displaystyle= dαd+ddααdd+dαd𝑑𝛼superscript𝑑𝑑superscript𝑑𝛼𝛼superscript𝑑𝑑superscript𝑑𝛼𝑑\displaystyle-d\alpha d^{{\dagger}}+dd^{{\dagger}}\alpha-\alpha d^{{\dagger}}d+d^{{\dagger}}\alpha d

After some computation where one uses cyclicity of trace and the fact that d𝑑d and dsuperscript𝑑d^{{\dagger}} commute with \triangle, it follows that

dduk=0d(1)kk2trΩketk𝑑𝑑𝑢superscriptsubscript𝑘0𝑑superscript1𝑘𝑘2subscripttrsubscriptΩ𝑘superscript𝑒𝑡subscript𝑘\displaystyle\frac{d}{du}\sum_{k=0}^{d}-\frac{(-1)^{k}k}{2}{\mbox{tr}}_{\Omega_{k}}e^{-t\triangle_{k}} =\displaystyle= 12tddtk=0d(1)ktrΩk(etkα)12𝑡𝑑𝑑𝑡superscriptsubscript𝑘0𝑑superscript1𝑘subscripttrsubscriptΩ𝑘superscript𝑒𝑡subscript𝑘𝛼\displaystyle-\frac{1}{2}t\frac{d}{dt}\sum_{k=0}^{d}(-1)^{k}{\mbox{tr}}_{\Omega_{k}}\left(e^{-t\triangle_{k}}\alpha\right)

and further that

uF(s)𝑢𝐹𝑠\displaystyle\frac{\partial}{\partial u}F(s) =\displaystyle= 12k=0d(1)ksΓ(s)0𝑑tts1trΩk(etkα)12superscriptsubscript𝑘0𝑑superscript1𝑘𝑠Γ𝑠superscriptsubscript0differential-d𝑡superscript𝑡𝑠1subscripttrsubscriptΩ𝑘superscript𝑒𝑡subscript𝑘𝛼\displaystyle\frac{1}{2}\sum_{k=0}^{d}(-1)^{k}\frac{s}{\Gamma(s)}\int_{0}^{\infty}dtt^{s-1}{\mbox{tr}}_{\Omega_{k}}\left(e^{-t\triangle_{k}}\alpha\right)

For d𝑑d odd we have a theorem that says that the integral of this expression has no pole in s𝑠s at s=0𝑠0s=0. Therefore we have a double zero at s=0𝑠0s=0 since Γ(s)=1/s+regularΓ𝑠1𝑠𝑟𝑒𝑔𝑢𝑙𝑎𝑟\Gamma(s)=1/s+regular. Therefore

uF(0)𝑢superscript𝐹0\displaystyle\frac{\partial}{\partial u}F^{\prime}(0) =\displaystyle= 00\displaystyle 0

proving metric independence of the analytic torsion.

If there are zero modes, then we have to first project those out from the definition of the analytic torsion. Let P𝑃P be the projection from forms in k=0dΩksuperscriptsubscriptdirect-sum𝑘0𝑑subscriptΩ𝑘\oplus_{k=0}^{d}\Omega_{k} to harmonic forms in k=0dHarmksuperscriptsubscriptdirect-sum𝑘0𝑑𝐻𝑎𝑟subscript𝑚𝑘\oplus_{k=0}^{d}Harm_{k}. Then we replace trΩksubscripttrsubscriptΩ𝑘{\mbox{tr}}_{\Omega_{k}} above with trΩk(1P)subscripttrsubscriptΩ𝑘1𝑃{\mbox{tr}}_{\Omega_{k}}(1-P). Formal manipulations now yield a nonzero contribution from the metric variation of the analytic torsion which is

uF(s)𝑢𝐹𝑠\displaystyle\frac{\partial}{\partial u}F(s) =\displaystyle= 12k=0d(1)ksΓ(s)0𝑑tts1trHarmk(α)12superscriptsubscript𝑘0𝑑superscript1𝑘𝑠Γ𝑠superscriptsubscript0differential-d𝑡superscript𝑡𝑠1subscripttr𝐻𝑎𝑟subscript𝑚𝑘𝛼\displaystyle-\frac{1}{2}\sum_{k=0}^{d}(-1)^{k}\frac{s}{\Gamma(s)}\int_{0}^{\infty}dtt^{s-1}{\mbox{tr}}_{Harm_{k}}\left(\alpha\right)

The integral is now elementary, and formally we have

0𝑑tts1superscriptsubscript0differential-d𝑡superscript𝑡𝑠1\displaystyle\int_{0}^{\infty}dtt^{s-1} =\displaystyle= 1s1𝑠\displaystyle-\frac{1}{s}

What we really do here is to compute the integral in the domain of s𝑠s where it is convergent and then we continue that result analytically in s𝑠s. We then have

uF(s)𝑢𝐹𝑠\displaystyle\frac{\partial}{\partial u}F(s) =\displaystyle= 12k=0d(1)k1Γ(s)trHarmk(α)12superscriptsubscript𝑘0𝑑superscript1𝑘1Γ𝑠subscripttr𝐻𝑎𝑟subscript𝑚𝑘𝛼\displaystyle-\frac{1}{2}\sum_{k=0}^{d}(-1)^{k}\frac{1}{\Gamma(s)}{\mbox{tr}}_{Harm_{k}}\left(\alpha\right)

By noting that Γ(s)=1/s+regΓ𝑠1𝑠𝑟𝑒𝑔\Gamma(s)=1/s+reg we see that

uF(0)𝑢superscript𝐹0\displaystyle\frac{\partial}{\partial u}F^{\prime}(0) =\displaystyle= 12k=0d(1)ktrHarmk(α)12superscriptsubscript𝑘0𝑑superscript1𝑘subscripttr𝐻𝑎𝑟subscript𝑚𝑘𝛼\displaystyle-\frac{1}{2}\sum_{k=0}^{d}(-1)^{k}{\mbox{tr}}_{Harm_{k}}\left(\alpha\right)

and so now we have

ulogτosc=uF(0)=12k=0d(1)ktrHarmk(α)𝑢subscript𝜏𝑜𝑠𝑐𝑢superscript𝐹012superscriptsubscript𝑘0𝑑superscript1𝑘subscripttr𝐻𝑎𝑟subscript𝑚𝑘𝛼\displaystyle\frac{\partial}{\partial u}\log\tau_{osc}=-\frac{\partial}{\partial u}F^{\prime}(0)=\frac{1}{2}\sum_{k=0}^{d}(-1)^{k}{\mbox{tr}}_{Harm_{k}}\left(\alpha\right)

which is nonzero. We then need to add a zero mode contribution whose variation cancels the above variation to get a metric-independent Ray-Singer torsion. Let us define

logτzerosubscript𝜏𝑧𝑒𝑟𝑜\displaystyle\log\tau_{zero} =\displaystyle= 12k=0di=1bk(M)(1)k(ωik,ωik)u12superscriptsubscript𝑘0𝑑superscriptsubscript𝑖1subscript𝑏𝑘𝑀superscript1𝑘subscriptsubscriptsuperscript𝜔𝑘𝑖subscriptsuperscript𝜔𝑘𝑖𝑢\displaystyle-\frac{1}{2}\sum_{k=0}^{d}\sum_{i=1}^{b_{k}(M)}(-1)^{k}(\omega^{k}_{i},\omega^{k}_{i})_{u}

where ωiksubscriptsuperscript𝜔𝑘𝑖\omega^{k}_{i} is a metric-independent and orthonormal basis of Harmonic k𝑘k-forms at u=0𝑢0u=0,

(ωik,ωjk)u=0subscriptsubscriptsuperscript𝜔𝑘𝑖subscriptsuperscript𝜔𝑘𝑗𝑢0\displaystyle(\omega^{k}_{i},\omega^{k}_{j})_{u=0} =\displaystyle= δijsubscript𝛿𝑖𝑗\displaystyle\delta_{ij}

where we define

(ω,ω)usubscript𝜔𝜔𝑢\displaystyle(\omega,\omega)_{u} =\displaystyle= Mωuω\displaystyle\int_{M}\omega\wedge*_{u}\omega

Then

u(ω,ω)u𝑢subscript𝜔𝜔𝑢\displaystyle\frac{\partial}{\partial u}(\omega,\omega)_{u} =\displaystyle= Mω˙ωsubscript𝑀𝜔˙𝜔\displaystyle\int_{M}\omega\wedge\dot{*}\omega
=\displaystyle= Mω˙ω\displaystyle\int_{M}\omega\wedge**\dot{*}\omega
=\displaystyle= Mωαω\displaystyle\int_{M}\omega\wedge*\alpha\omega

and so we find its metric variation evaluated at u=0𝑢0u=0 as

ulogτzero=12k=0di=1bk(M)(1)ktrHarmk(α)𝑢subscript𝜏𝑧𝑒𝑟𝑜12superscriptsubscript𝑘0𝑑superscriptsubscript𝑖1subscript𝑏𝑘𝑀superscript1𝑘subscripttr𝐻𝑎𝑟subscript𝑚𝑘𝛼\displaystyle\frac{\partial}{\partial u}\log\tau_{zero}=-\frac{1}{2}\sum_{k=0}^{d}\sum_{i=1}^{b_{k}(M)}(-1)^{k}{\mbox{tr}}_{Harm_{k}}\left(\alpha\right)

which precisely cancels the metric variation of the analytic torsion. This completes the proof.

However, τzerosubscript𝜏𝑧𝑒𝑟𝑜\tau_{zero} is not uniquely fixed by the requirement that ωiksubscriptsuperscript𝜔𝑘𝑖\omega^{k}_{i} form an orthonormal basis at u=0𝑢0u=0, since there are many ways that we can introduce a parameter u𝑢u and the metric is not uniquely fixed by the condition u=0𝑢0u=0. To improve this situation, we will fix the point u=0𝑢0u=0 by the requirement

Mu=01subscript𝑢0subscript𝑀1\displaystyle\int_{M}*_{u=0}1 =\displaystyle= 11\displaystyle 1 (A.4)

A.1 Explicit computations

Let us compute the Ray-Singer torsion for a circle with the metric ds2=r2dθ2𝑑superscript𝑠2superscript𝑟2𝑑superscript𝜃2ds^{2}=r^{2}d\theta^{2}. The Hodge operator acts as 1=rdθ*1=rd\theta and dθ=1r*d\theta=\frac{1}{r}. We have

τoscsubscript𝜏𝑜𝑠𝑐\displaystyle\tau_{osc} =\displaystyle= n0(nr)2subscriptproduct𝑛0superscript𝑛𝑟2\displaystyle\sqrt{\prod_{n\neq 0}\left(\frac{n}{r}\right)^{2}}

which can be computed using zeta function regularization with the result

τoscsubscript𝜏𝑜𝑠𝑐\displaystyle\tau_{osc} =\displaystyle= 2πr2𝜋𝑟\displaystyle 2\pi r

Note that τoscsubscript𝜏𝑜𝑠𝑐\tau_{osc} depends on the metric. We need to multiply by τzerosubscript𝜏𝑧𝑒𝑟𝑜\tau_{zero} to get a metric-independent result. On S1superscript𝑆1S^{1} there are 0-form and 1-form zero modes

ω0subscript𝜔0\displaystyle\omega_{0} =\displaystyle= 11\displaystyle 1
ω1subscript𝜔1\displaystyle\omega_{1} =\displaystyle= dθ2π𝑑𝜃2𝜋\displaystyle\frac{d\theta}{2\pi}

These are chosen such that they are orthonormal at r0=12πsubscript𝑟012𝜋r_{0}=\frac{1}{2\pi} where the circumference is one, 2πr0=12𝜋subscript𝑟012\pi r_{0}=1, and hence corresponds to the point u=0𝑢0u=0. For example, we could let r=12π+u𝑟12𝜋𝑢r=\frac{1}{2\pi}+u. For a generic radius r𝑟r, we get

(ω0,ω0)subscript𝜔0subscript𝜔0\displaystyle(\omega_{0},\omega_{0}) =\displaystyle= 2πr2𝜋𝑟\displaystyle 2\pi r
(ω1,ω1)subscript𝜔1subscript𝜔1\displaystyle(\omega_{1},\omega_{1}) =\displaystyle= 12πr12𝜋𝑟\displaystyle\frac{1}{2\pi r}

Then

τzerosubscript𝜏𝑧𝑒𝑟𝑜\displaystyle\tau_{zero} =\displaystyle= 12πr12𝜋𝑟\displaystyle\frac{1}{2\pi r}

and we get

τ=τzeroτosc=1𝜏subscript𝜏𝑧𝑒𝑟𝑜subscript𝜏𝑜𝑠𝑐1\displaystyle\tau=\tau_{zero}\tau_{osc}=1

which is independent of the metric.

Let us now compute the Ray-Singer torsion for S5superscript𝑆5S^{5}. First we compute

τoscsubscript𝜏𝑜𝑠𝑐\displaystyle\tau_{osc} =\displaystyle= det0coexdet122coexdet1coexsuperscriptsuperscriptsubscript0𝑐𝑜𝑒𝑥superscript12superscriptsubscript2𝑐𝑜𝑒𝑥superscriptsuperscriptsubscript1𝑐𝑜𝑒𝑥\displaystyle\frac{\det^{\prime}\triangle_{0}^{coex}\det^{\prime\frac{1}{2}}\triangle_{2}^{coex}}{\det^{\prime}\triangle_{1}^{coex}}

We can compute this knowing the eigenvalues,

(λ0)nsubscriptsubscript𝜆0𝑛\displaystyle(\lambda_{0})_{n} =\displaystyle= 1r2n(n+4)1superscript𝑟2𝑛𝑛4\displaystyle\frac{1}{r^{2}}n(n+4)
(λ1)nsubscriptsubscript𝜆1𝑛\displaystyle(\lambda_{1})_{n} =\displaystyle= 1r2(n+1)(n+3)1superscript𝑟2𝑛1𝑛3\displaystyle\frac{1}{r^{2}}(n+1)(n+3)
(λ2+)nsubscriptsuperscriptsubscript𝜆2𝑛\displaystyle(\lambda_{2}^{+})_{n} =\displaystyle= 1r2(n+2)21superscript𝑟2superscript𝑛22\displaystyle\frac{1}{r^{2}}(n+2)^{2}

and the degeneracies

(b0)nsubscriptsubscript𝑏0𝑛\displaystyle(b_{0})_{n} =\displaystyle= 112(n+1)(n+2)2(n+3)112𝑛1superscript𝑛22𝑛3\displaystyle\frac{1}{12}(n+1)(n+2)^{2}(n+3)
(b1)nsubscriptsubscript𝑏1𝑛\displaystyle(b_{1})_{n} =\displaystyle= 13n(n+2)2(n+4)13𝑛superscript𝑛22𝑛4\displaystyle\frac{1}{3}n(n+2)^{2}(n+4)
(b2+)nsubscriptsuperscriptsubscript𝑏2𝑛\displaystyle(b_{2}^{+})_{n} =\displaystyle= 14n(n+1)(n+3)(n+4)14𝑛𝑛1𝑛3𝑛4\displaystyle\frac{1}{4}n(n+1)(n+3)(n+4)

of the spherical harmonics. Here n=0,1,2,𝑛012n=0,1,2,... and we see that we have one zero eigenvalue (λ0)0=0subscriptsubscript𝜆000(\lambda_{0})_{0}=0 with degeneracy (b0)0=1subscriptsubscript𝑏001(b_{0})_{0}=1. We will take out this zero mode. Furthermore, since (b1)0=0subscriptsubscript𝑏100(b_{1})_{0}=0 and (b2+)0=0subscriptsuperscriptsubscript𝑏200(b_{2}^{+})_{0}=0, the contribution from the remaining modes with n=0𝑛0n=0 gives simply a multiplicative factor of 111 that we can forget about. Then the rest becomes

lnτoscsubscript𝜏𝑜𝑠𝑐\displaystyle\ln\tau_{osc} =\displaystyle= n=1[(b0)n(lnnr+lnn+4r)+(b2+)n2lnn+2r(b1)n(lnn+3r+lnn+1r)]superscriptsubscript𝑛1delimited-[]subscriptsubscript𝑏0𝑛𝑛𝑟𝑛4𝑟subscriptsubscriptsuperscript𝑏2𝑛2𝑛2𝑟subscriptsubscript𝑏1𝑛𝑛3𝑟𝑛1𝑟\displaystyle\sum_{n=1}^{\infty}\left[(b_{0})_{n}\left(\ln\frac{n}{r}+\ln\frac{n+4}{r}\right)+(b^{+}_{2})_{n}2\ln\frac{n+2}{r}-(b_{1})_{n}\left(\ln\frac{n+3}{r}+\ln\frac{n+1}{r}\right)\right]

Before we apply zeta function regularization, we shift n𝑛n such that we get the sum in the form

lnτoscsubscript𝜏𝑜𝑠𝑐\displaystyle\ln\tau_{osc} =\displaystyle= n=1((b0)n+4+(b0)n+2(b2+)n+2(b1)n+1(b1)n+3)lnn+4rsuperscriptsubscript𝑛1subscriptsubscript𝑏0𝑛4subscriptsubscript𝑏0𝑛2subscriptsuperscriptsubscript𝑏2𝑛2subscriptsubscript𝑏1𝑛1subscriptsubscript𝑏1𝑛3𝑛4𝑟\displaystyle\sum_{n=1}^{\infty}\left((b_{0})_{n+4}+(b_{0})_{n}+2(b_{2}^{+})_{n+2}-(b_{1})_{n+1}-(b_{1})_{n+3}\right)\ln\frac{n+4}{r}
+(b0)1ln1r+(b0)2ln2r+(b0)3ln3r+(b0)4ln4rsubscriptsubscript𝑏011𝑟subscriptsubscript𝑏022𝑟subscriptsubscript𝑏033𝑟subscriptsubscript𝑏044𝑟\displaystyle+(b_{0})_{1}\ln\frac{1}{r}+(b_{0})_{2}\ln\frac{2}{r}+(b_{0})_{3}\ln\frac{3}{r}+(b_{0})_{4}\ln\frac{4}{r}
+(b2)12ln3r+(b2)22ln4rsubscriptsubscript𝑏2123𝑟subscriptsubscript𝑏2224𝑟\displaystyle+(b_{2})_{1}2\ln\frac{3}{r}+(b_{2})_{2}2\ln\frac{4}{r}
(b1)1ln4r(b1)1ln2r(b1)2ln3r(b1)3ln4rsubscriptsubscript𝑏114𝑟subscriptsubscript𝑏112𝑟subscriptsubscript𝑏123𝑟subscriptsubscript𝑏134𝑟\displaystyle-(b_{1})_{1}\ln\frac{4}{r}-(b_{1})_{1}\ln\frac{2}{r}-(b_{1})_{2}\ln\frac{3}{r}-(b_{1})_{3}\ln\frac{4}{r}

Using Mathematica we find that this simplifies to

lnτoscsubscript𝜏𝑜𝑠𝑐\displaystyle\ln\tau_{osc} =\displaystyle= n=16lnn+4rsuperscriptsubscript𝑛16𝑛4𝑟\displaystyle\sum_{n=1}^{\infty}6\ln\frac{n+4}{r}
+5ln2r+6ln3r+5ln4r52𝑟63𝑟54𝑟\displaystyle+5\ln\frac{2}{r}+6\ln\frac{3}{r}+5\ln\frac{4}{r}
=\displaystyle= 6(n=1lnnr)ln8r26superscriptsubscript𝑛1𝑛𝑟8superscript𝑟2\displaystyle 6\left(\sum_{n=1}^{\infty}\ln\frac{n}{r}\right)-\ln\frac{8}{r^{2}}

Now we apply zeta function regularization on the infinite sum, and get

τoscsubscript𝜏𝑜𝑠𝑐\displaystyle\tau_{osc} =\displaystyle= π3r5superscript𝜋3superscript𝑟5\displaystyle\pi^{3}r^{5}

We notice that this is the volume of S5superscript𝑆5S^{5}. We shall now choose normalization for our harmonic forms on S5superscript𝑆5S^{5}. These are

ω0subscript𝜔0\displaystyle\omega_{0} =\displaystyle= 11\displaystyle 1
ω5subscript𝜔5\displaystyle\omega_{5} =\displaystyle= Ω^5π3subscript^Ω5superscript𝜋3\displaystyle\frac{\widehat{\Omega}_{5}}{\pi^{3}}

where Ω^5subscript^Ω5\widehat{\Omega}_{5} denotes the volume-form of the unit five-sphere. Then

(ω0,ω0)subscript𝜔0subscript𝜔0\displaystyle(\omega_{0},\omega_{0}) =\displaystyle= π3r5superscript𝜋3superscript𝑟5\displaystyle\pi^{3}r^{5}
(ω5,ω5)subscript𝜔5subscript𝜔5\displaystyle(\omega_{5},\omega_{5}) =\displaystyle= 1π3r51superscript𝜋3superscript𝑟5\displaystyle\frac{1}{\pi^{3}r^{5}}

and we get

τzero=(ω0,ω0)12(ω5,ω5)12=1π3r5subscript𝜏𝑧𝑒𝑟𝑜superscriptsubscript𝜔0subscript𝜔012superscriptsubscript𝜔5subscript𝜔5121superscript𝜋3superscript𝑟5\displaystyle\tau_{zero}=(\omega_{0},\omega_{0})^{-\frac{1}{2}}(\omega_{5},\omega_{5})^{\frac{1}{2}}=\frac{1}{\pi^{3}r^{5}}

and so we get

τ𝜏\displaystyle\tau =\displaystyle= 11\displaystyle 1

For S1/psuperscript𝑆1subscript𝑝S^{1}/\mathbb{Z}_{p} which is again a circle, we have shown that we have τ=1𝜏1\tau=1.

Let us now proceed to S5/p=L(p;1,1)superscript𝑆5subscript𝑝𝐿𝑝11S^{5}/\mathbb{Z}_{p}=L(p;1,1). For the zero mode part, as our harmonic p𝑝p-forms, we take

ω0subscript𝜔0\displaystyle\omega_{0} =\displaystyle= 11\displaystyle 1
ω5subscript𝜔5\displaystyle\omega_{5} =\displaystyle= pΩ^5π3𝑝subscript^Ω5superscript𝜋3\displaystyle\frac{p\widehat{\Omega}_{5}}{\pi^{3}}

Then

(ω0,ω0)subscript𝜔0subscript𝜔0\displaystyle(\omega_{0},\omega_{0}) =\displaystyle= π3r5psuperscript𝜋3superscript𝑟5𝑝\displaystyle\frac{\pi^{3}r^{5}}{p}
(ω5,ω5)subscript𝜔5subscript𝜔5\displaystyle(\omega_{5},\omega_{5}) =\displaystyle= pπ3r5𝑝superscript𝜋3superscript𝑟5\displaystyle\frac{p}{\pi^{3}r^{5}}

From this we get

τzerosubscript𝜏𝑧𝑒𝑟𝑜\displaystyle\tau_{zero} =\displaystyle= pπ3r5𝑝superscript𝜋3superscript𝑟5\displaystyle\frac{p}{\pi^{3}r^{5}}

Let us now turn to the oscillator modes (and let us temporarily put the radius r=1𝑟1r=1). We define

d(p,q)𝑑𝑝𝑞\displaystyle d(p,q) =\displaystyle= 12(p+1)(q+1)(p+q+2)12𝑝1𝑞1𝑝𝑞2\displaystyle\frac{1}{2}(p+1)(q+1)(p+q+2)

as the dimension of the representation labeled (p,q)𝑝𝑞(p,q) of SU(3)𝑆𝑈3SU(3). We then introduce the following refined dimensions of representations of SU(4)=SO(6)𝑆𝑈4𝑆𝑂6SU(4)=SO(6) [32],

d(n,a,0)𝑑𝑛𝑎0\displaystyle d(n,a,0) =\displaystyle= k=0nd(k,nk)eia(2kn)superscriptsubscript𝑘0𝑛𝑑𝑘𝑛𝑘superscript𝑒𝑖𝑎2𝑘𝑛\displaystyle\sum_{k=0}^{n}d(k,n-k)e^{ia(2k-n)}
d(n,a,1)𝑑𝑛𝑎1\displaystyle d(n,a,1) =\displaystyle= k=0n1[(d(k,nk1)eia(2kn+1)+d(k1,nk1)eia(2kn1)\displaystyle\sum_{k=0}^{n-1}\Big{[}(d(k,n-k-1)e^{ia(2k-n+1)}+d(k-1,n-k-1)e^{ia(2k-n-1)}
+d(k+1,nk)eia(2kn+1)+d(k,nk)eia(2kn+3)]\displaystyle+d(k+1,n-k)e^{ia(2k-n+1)}+d(k,n-k)e^{ia(2k-n+3)}\Big{]}
d(n,a,2)𝑑𝑛𝑎2\displaystyle d(n,a,2) =\displaystyle= k=0n1[d(k,nk1)eia(2kn2)+d(k,nk)eia(2kn)+d(k,nk+1)eia(2kn+2)]superscriptsubscript𝑘0𝑛1delimited-[]𝑑𝑘𝑛𝑘1superscript𝑒𝑖𝑎2𝑘𝑛2𝑑𝑘𝑛𝑘superscript𝑒𝑖𝑎2𝑘𝑛𝑑𝑘𝑛𝑘1superscript𝑒𝑖𝑎2𝑘𝑛2\displaystyle\sum_{k=0}^{n-1}\Big{[}d(k,n-k-1)e^{ia(2k-n-2)}+d(k,n-k)e^{ia(2k-n)}+d(k,n-k+1)e^{ia(2k-n+2)}\Big{]}

We then define

d(n,p,rank)𝑑𝑛𝑝𝑟𝑎𝑛𝑘\displaystyle d(n,p,rank) =\displaystyle= 1p=0p1d(n,2πp,rank)1𝑝superscriptsubscript0𝑝1𝑑𝑛2𝜋𝑝𝑟𝑎𝑛𝑘\displaystyle\frac{1}{p}\sum_{{{\ell}}=0}^{p-1}d\left(n,\frac{2\pi{{\ell}}}{p},rank\right)

For p=1𝑝1p=1 these are

d(n,1,0)𝑑𝑛10\displaystyle d(n,1,0) =\displaystyle= (b0)nsubscriptsubscript𝑏0𝑛\displaystyle(b_{0})_{n}
d(n,1,1)𝑑𝑛11\displaystyle d(n,1,1) =\displaystyle= (b1)nsubscriptsubscript𝑏1𝑛\displaystyle(b_{1})_{n}
d(n,1,2)𝑑𝑛12\displaystyle d(n,1,2) =\displaystyle= (b2+)nsubscriptsuperscriptsubscript𝑏2𝑛\displaystyle(b_{2}^{+})_{n}

and for p=2𝑝2p=2 they are

d(n,2,0)𝑑𝑛20\displaystyle d(n,2,0) =\displaystyle= 12(1+(1)n)(b0)n121superscript1𝑛subscriptsubscript𝑏0𝑛\displaystyle\frac{1}{2}\left(1+(-1)^{n}\right)(b_{0})_{n}
d(n,2,1)𝑑𝑛21\displaystyle d(n,2,1) =\displaystyle= 12(1(1)n)(b1)n121superscript1𝑛subscriptsubscript𝑏1𝑛\displaystyle\frac{1}{2}\left(1-(-1)^{n}\right)(b_{1})_{n}
d(n,2,2)𝑑𝑛22\displaystyle d(n,2,2) =\displaystyle= 12(1+(1)n)(b2+)n121superscript1𝑛subscriptsuperscriptsubscript𝑏2𝑛\displaystyle\frac{1}{2}\left(1+(-1)^{n}\right)(b_{2}^{+})_{n}

and for higher values of p𝑝p we may obtain corresponding, but much more complicated, expressions for the dimensions of the representations for spherical harmonics on S5/psuperscript𝑆5subscript𝑝S^{5}/\mathbb{Z}_{p}. We notice that the result for p=2𝑝2p=2 reflects the fact that we keep those spherical harmonics which are even under zizisubscript𝑧𝑖subscript𝑧𝑖z_{i}\rightarrow-z_{i} if we embed S5superscript𝑆5S^{5} into 3superscript3\mathbb{C}^{3} with complex coordinates zisubscript𝑧𝑖z_{i} (i=1,2,3𝑖123i=1,2,3). For the scalar and two-form, these are spherical harmonics of even degree, while for the vector spherical harmonics, those are of odd degree n𝑛n.

We then define

D(n,p)𝐷𝑛𝑝\displaystyle D(n,p) =\displaystyle= d(n+4,p,0)+d(n,p,0)+2d(n+2,p,2)d(n+1,p,1)d(n+3,p,1)𝑑𝑛4𝑝0𝑑𝑛𝑝02𝑑𝑛2𝑝2𝑑𝑛1𝑝1𝑑𝑛3𝑝1\displaystyle d(n+4,p,0)+d(n,p,0)+2d(n+2,p,2)-d(n+1,p,1)-d(n+3,p,1)

and

D(p)𝐷𝑝\displaystyle D(p) =\displaystyle= d(2,p,0)ln(2)+d(3,p,0)ln(3)+d(4,p,0)ln(4)𝑑2𝑝02𝑑3𝑝03𝑑4𝑝04\displaystyle d(2,p,0)\ln(2)+d(3,p,0)\ln(3)+d(4,p,0)\ln(4)
+2d(1,p,2)ln(3)+2d(2,p,2)ln(4)2𝑑1𝑝232𝑑2𝑝24\displaystyle+2d(1,p,2)\ln(3)+2d(2,p,2)\ln(4)
d(1,p,1)ln(4)d(1,p,1)ln(2)d(2,p,1)ln(3)d(3,p,1)ln(4)𝑑1𝑝14𝑑1𝑝12𝑑2𝑝13𝑑3𝑝14\displaystyle-d(1,p,1)\ln(4)-d(1,p,1)\ln(2)-d(2,p,1)\ln(3)-d(3,p,1)\ln(4)

We computed these quantities up to p=4𝑝4p=4 with Mathematica. If we define ωp:=exp2πpassignsubscript𝜔𝑝2𝜋𝑝\omega_{p}:=\exp\frac{2\pi}{p}, then we can express the results as

D(n,1)𝐷𝑛1\displaystyle D(n,1) =\displaystyle= 66\displaystyle 6
D(n,2)𝐷𝑛2\displaystyle D(n,2) =\displaystyle= 612(1+ω2n)6121superscriptsubscript𝜔2𝑛\displaystyle 6\frac{1}{2}\left(1+\omega_{2}^{n}\right)
D(n,3)𝐷𝑛3\displaystyle D(n,3) =\displaystyle= 613(1+ω3n2+ω32(n2))6131superscriptsubscript𝜔3𝑛2superscriptsubscript𝜔32𝑛2\displaystyle 6\frac{1}{3}\left(1+\omega_{3}^{n-2}+\omega_{3}^{2(n-2)}\right)
D(n,4)𝐷𝑛4\displaystyle D(n,4) =\displaystyle= 614(1+ω4n+ω42n+ω43n)6141superscriptsubscript𝜔4𝑛superscriptsubscript𝜔42𝑛superscriptsubscript𝜔43𝑛\displaystyle 6\frac{1}{4}\left(1+\omega_{4}^{n}+\omega_{4}^{2n}+\omega_{4}^{3n}\right)

and

D(1)𝐷1\displaystyle D(1) =\displaystyle= 5ln(2)+6ln(3)+5ln(4)526354\displaystyle 5\ln(2)+6\ln(3)+5\ln(4)
D(2)𝐷2\displaystyle D(2) =\displaystyle= 5ln(2)+5ln(4)5254\displaystyle 5\ln(2)+5\ln(4)
D(3)𝐷3\displaystyle D(3) =\displaystyle= ln(2)+6ln(3)ln(4)2634\displaystyle-\ln(2)+6\ln(3)-\ln(4)
D(4)𝐷4\displaystyle D(4) =\displaystyle= ln(2)+5ln(4)254\displaystyle-\ln(2)+5\ln(4)

Also, for n4𝑛4n\geq 4, we get

D(n)𝐷𝑛\displaystyle D(n) =\displaystyle= ln(2)ln(4)24\displaystyle-\ln(2)-\ln(4)

Putting these results together, we find that, at least up to p=4𝑝4p=4,

lnτoscsubscript𝜏𝑜𝑠𝑐\displaystyle\ln\tau_{osc} =\displaystyle= ln(8)+6k=1ln(pk)86superscriptsubscript𝑘1𝑝𝑘\displaystyle-\ln(8)+6\sum_{k=1}^{\infty}\ln(pk)

which by zeta function regularization leads to

τoscsubscript𝜏𝑜𝑠𝑐\displaystyle\tau_{osc} =\displaystyle= π3r5p3superscript𝜋3superscript𝑟5superscript𝑝3\displaystyle\frac{\pi^{3}r^{5}}{p^{3}}

We believe this formula is valid for any positive integers p𝑝p although we have checked it only for p=1,2,3,4𝑝1234p=1,2,3,4. Combining this with the zero mode contribution τzerosubscript𝜏𝑧𝑒𝑟𝑜\tau_{zero}, we get

τ𝜏\displaystyle\tau =\displaystyle= 1p21superscript𝑝2\displaystyle\frac{1}{p^{2}}

Our results on S1/psuperscript𝑆1subscript𝑝S^{1}/\mathbb{Z}_{p} and S5/psuperscript𝑆5subscript𝑝S^{5}/\mathbb{Z}_{p} are now consistent with a general formula for the Ray-Singer torsion on S2N1/psuperscript𝑆2𝑁1subscript𝑝S^{2N-1}/\mathbb{Z}_{p},

τ𝜏\displaystyle\tau =\displaystyle= 1pN11superscript𝑝𝑁1\displaystyle\frac{1}{p^{N-1}}

which we stated as a conjecture in the main text as Eq. (2.3).

Appendix B The Minakshisundaram-Pleijel theorem

The Minakshisundaram-Pleijel theorem [33] says that when d𝑑d is odd, the number of zero modes of the Laplacian acting on the space of p𝑝p-forms, is encoded in the spectrum of the non-harmonic forms,

ζp(0)subscript𝜁subscript𝑝0\displaystyle\zeta_{\triangle_{p}}(0) =\displaystyle= bpsubscript𝑏𝑝\displaystyle-b_{p} (B.1)

where bp=dimKerpsubscript𝑏𝑝dimensionKersubscript𝑝b_{p}=\dim{\mbox{Ker}}\triangle_{p}. The regularized value for the determinant of psubscript𝑝\triangle_{p} is given by

det(p)subscript𝑝\displaystyle\det\left(\triangle_{p}\right) =\displaystyle= eζp(0)superscript𝑒superscriptsubscript𝜁subscript𝑝0\displaystyle e^{-\zeta_{\triangle_{p}}^{\prime}(0)}

By noting that for C𝐶C\in\mathbb{C}

ζCp(s)superscriptsubscript𝜁𝐶subscript𝑝𝑠\displaystyle\zeta_{C\triangle_{p}}^{\prime}(s) =\displaystyle= (ζp(s)logC+ζp(s))Cssubscript𝜁subscript𝑝𝑠𝐶superscriptsubscript𝜁subscript𝑝𝑠superscript𝐶𝑠\displaystyle\left(\zeta_{\triangle_{p}}(s)\log C+\zeta_{\triangle_{p}}^{\prime}(s)\right)C^{-s}

we get

det(Cp)=Cζp(0)det(p)=Cbpdet(p)𝐶subscript𝑝superscript𝐶subscript𝜁subscript𝑝0subscript𝑝superscript𝐶subscript𝑏𝑝subscript𝑝\displaystyle\det\left(C\triangle_{p}\right)=C^{\zeta_{\triangle_{p}}(0)}\det\left(\triangle_{p}\right)=C^{-b_{p}}\det\left(\triangle_{p}\right) (B.2)

Appendix C Partial gauge fixing by the Faddeev-Popov method

Here we illustrate a general theorem in [11] by a very simple example that we borrow from the appendix in [8]. We consider the ‘path integral’ in zero dimensions with target space 2superscript2\mathbb{R}^{2},

Z𝑍\displaystyle Z =\displaystyle= 1Vol(G)2dX2πdY2πeKS(R)1Vol𝐺subscriptsuperscript2𝑑𝑋2𝜋𝑑𝑌2𝜋superscript𝑒𝐾𝑆𝑅\displaystyle\frac{1}{{\mbox{Vol}}(G)}\int_{\mathbb{R}^{2}}\frac{dX}{\sqrt{2\pi}}\frac{dY}{\sqrt{2\pi}}e^{-KS(R)}

where G=U(1)𝐺𝑈1G=U(1) is a gauge symmetry and Vol(G)=2πVol𝐺2𝜋{\mbox{Vol}}(G)=2\pi is its volume. If we assume that R0>0subscript𝑅00R_{0}>0 is a minimum for the ‘action’ S(R)𝑆𝑅S(R), then the saddle point approximation gives

Z𝑍\displaystyle Z =\displaystyle= R0eKS(R0)2πKS′′(R0)subscript𝑅0superscript𝑒𝐾𝑆subscript𝑅02𝜋𝐾superscript𝑆′′subscript𝑅0\displaystyle\frac{R_{0}e^{-KS(R_{0})}}{\sqrt{2\pi KS^{\prime\prime}(R_{0})}}

which is a good approximation when K𝐾K is large.

On the other hand, we can use the U(1)𝑈1U(1) gauge symmetry that acts on the ‘fields’ as

(XΛYΛ)matrixsuperscript𝑋Λsuperscript𝑌Λ\displaystyle\begin{pmatrix}X^{\Lambda}\\ Y^{\Lambda}\end{pmatrix} =\displaystyle= (cosΛsinΛsinΛcosΛ)(XY)matrixΛΛΛΛmatrix𝑋𝑌\displaystyle\begin{pmatrix}\cos\Lambda&\sin\Lambda\\ -\sin\Lambda&\cos\Lambda\end{pmatrix}\begin{pmatrix}X\\ Y\end{pmatrix}

to fix the gauge X=0𝑋0X=0. By the Faddeev-Popov procedure, we begin by defining a gauge fixing function

GΛsuperscript𝐺Λ\displaystyle G^{\Lambda} =\displaystyle= KR0XΛ𝐾subscript𝑅0superscript𝑋Λ\displaystyle KR_{0}X^{\Lambda}

From

1=𝑑GΛδ(GΛ)=𝑑ΛdGΛdΛδ(GΛ)=KR0𝑑ΛYΛδ(GΛ)=KR02𝑑Λδ(GΛ)1differential-dsuperscript𝐺Λ𝛿superscript𝐺Λdifferential-dΛ𝑑superscript𝐺Λ𝑑Λ𝛿superscript𝐺Λ𝐾subscript𝑅0differential-dΛsuperscript𝑌Λ𝛿superscript𝐺Λ𝐾superscriptsubscript𝑅02differential-dΛ𝛿superscript𝐺Λ\displaystyle 1=\int dG^{\Lambda}\delta(G^{\Lambda})=\int d\Lambda\frac{dG^{\Lambda}}{d\Lambda}\delta(G^{\Lambda})=KR_{0}\int d\Lambda Y^{\Lambda}\delta(G^{\Lambda})=KR_{0}^{2}\int d\Lambda\delta(G^{\Lambda})

we read off the FP determinant

detFPsubscript𝐹𝑃\displaystyle\det\triangle_{FP} =\displaystyle= KR02𝐾superscriptsubscript𝑅02\displaystyle KR_{0}^{2}

Inserting 111 into the path integral, we get

Z=KR02dX2πdY2πeKS(R)δ(KR0X)𝑍𝐾superscriptsubscript𝑅02𝑑𝑋2𝜋𝑑𝑌2𝜋superscript𝑒𝐾𝑆𝑅𝛿𝐾subscript𝑅0𝑋\displaystyle Z=KR_{0}^{2}\int\frac{dX}{\sqrt{2\pi}}\frac{dY}{\sqrt{2\pi}}e^{-KS(R)}\delta(KR_{0}X)

We write

δ(KR0X)=𝑑Ze2πiKR0ZX𝛿𝐾subscript𝑅0𝑋differential-d𝑍superscript𝑒2𝜋𝑖𝐾subscript𝑅0𝑍𝑋\displaystyle\delta(KR_{0}X)=\int dZe^{2\pi iKR_{0}ZX}

and expand about R0subscript𝑅0R_{0} to get

Z=KR02eKS(R0)dX2πdY2π𝑑Ze12KS′′(R0)(X2+Y2)e2πiKR0ZX𝑍𝐾superscriptsubscript𝑅02superscript𝑒𝐾𝑆subscript𝑅0𝑑𝑋2𝜋𝑑𝑌2𝜋differential-d𝑍superscript𝑒12𝐾superscript𝑆′′subscript𝑅0superscript𝑋2superscript𝑌2superscript𝑒2𝜋𝑖𝐾subscript𝑅0𝑍𝑋\displaystyle Z=KR_{0}^{2}e^{-KS(R_{0})}\int\frac{dX}{\sqrt{2\pi}}\frac{dY}{\sqrt{2\pi}}dZe^{-\frac{1}{2}KS^{\prime\prime}(R_{0})(X^{2}+Y^{2})}e^{2\pi iKR_{0}ZX}

The action involves the following matrix

L𝐿\displaystyle L =\displaystyle= K(S′′(R0)000S′′(R0)2πiR002πiR00)𝐾matrixsuperscript𝑆′′subscript𝑅0000superscript𝑆′′subscript𝑅02𝜋𝑖subscript𝑅002𝜋𝑖subscript𝑅00\displaystyle K\begin{pmatrix}S^{\prime\prime}(R_{0})&0&0\\ 0&S^{\prime\prime}(R_{0})&2\pi iR_{0}\\ 0&2\pi iR_{0}&0\end{pmatrix}

Then

Z=detFPdet(L)eKS(R0)=R0eKS(R0)2πKS′′(R0)𝑍subscript𝐹𝑃𝐿superscript𝑒𝐾𝑆subscript𝑅0subscript𝑅0superscript𝑒𝐾𝑆subscript𝑅02𝜋𝐾superscript𝑆′′subscript𝑅0\displaystyle Z=\frac{\det\triangle_{FP}}{\sqrt{\det(L)}}e^{-KS(R_{0})}=\frac{R_{0}e^{-KS(R_{0})}}{\sqrt{2\pi KS^{\prime\prime}(R_{0})}}

which agrees with the saddle point approximation.

Let us now instead assume that the global minimum of the action as at R0=0subscript𝑅00R_{0}=0. In this case

detFPsubscript𝐹𝑃\displaystyle\det\triangle_{FP} =\displaystyle= 00\displaystyle 0

and

L𝐿\displaystyle L =\displaystyle= K(S′′(R0)000S′′(R0)0000)𝐾matrixsuperscript𝑆′′subscript𝑅0000superscript𝑆′′subscript𝑅00000\displaystyle K\begin{pmatrix}S^{\prime\prime}(R_{0})&0&0\\ 0&S^{\prime\prime}(R_{0})&0\\ 0&0&0\end{pmatrix}

and the general formula

Z=detFPdet(L)eKS(R0)𝑍subscript𝐹𝑃𝐿superscript𝑒𝐾𝑆subscript𝑅0\displaystyle Z=\frac{\det\triangle_{FP}}{\sqrt{\det(L)}}e^{-KS(R_{0})}

becomes ill-defined since there is a fermionic ghost zero mode of FPsubscript𝐹𝑃\triangle_{FP} as well as a bosonic zero mode in L𝐿L. As a first try, we take out all those zero modes. Then we get

detFPsuperscriptsubscript𝐹𝑃\displaystyle\det{}^{\prime}\triangle_{FP} =\displaystyle= 11\displaystyle 1

(we shall define the determinant of an empty FP matrix to be 111 since that means we are not gauge fixing anything) and

Lsuperscript𝐿\displaystyle L^{\prime} =\displaystyle= K(S′′(R0)00S′′(R0))𝐾matrixsuperscript𝑆′′subscript𝑅000superscript𝑆′′subscript𝑅0\displaystyle K\begin{pmatrix}S^{\prime\prime}(R_{0})&0\\ 0&S^{\prime\prime}(R_{0})\\ \end{pmatrix}

and we would arrive at the result

Z=detFPdet(L)eKS(R0)\displaystyle Z=\frac{\det{}^{\prime}\triangle_{FP}}{\sqrt{\det{}^{\prime}(L)}}e^{-KS(R_{0})}

where primes indicate that the zero modes are taken out. It turns out that this gives almost the correct answer. We can compute Z𝑍Z for large K𝐾K without gauge fixing the U(1)𝑈1U(1) gauge symmetry at all,

Z𝑍\displaystyle Z =\displaystyle= 1Vol(G)eKS(0)dX2πdY2πeK2S′′(0)(X2+Y2)1Vol𝐺superscript𝑒𝐾𝑆0𝑑𝑋2𝜋𝑑𝑌2𝜋superscript𝑒𝐾2superscript𝑆′′0superscript𝑋2superscript𝑌2\displaystyle\frac{1}{{\mbox{Vol}}(G)}e^{-KS(0)}\int\frac{dX}{\sqrt{2\pi}}\frac{dY}{\sqrt{2\pi}}e^{-\frac{K}{2}S^{\prime\prime}(0)\left(X^{2}+Y^{2}\right)}

In this case the FP determinant is trivial,

detFPsubscript𝐹𝑃\displaystyle\det\triangle_{FP} =\displaystyle= 11\displaystyle 1

while the L𝐿L matrix is given by

L𝐿\displaystyle L =\displaystyle= K(S′′(0)00S′′(0))𝐾matrixsuperscript𝑆′′000superscript𝑆′′0\displaystyle K\begin{pmatrix}S^{\prime\prime}(0)&0\\ 0&S^{\prime\prime}(0)\end{pmatrix}

Then the result can be expressed as

Z𝑍\displaystyle Z =\displaystyle= 1Vol(G)detFPdet(L)eKS(0)1Vol𝐺subscript𝐹𝑃𝐿superscript𝑒𝐾𝑆0\displaystyle\frac{1}{{\mbox{Vol}}(G)}\frac{\det\triangle_{FP}}{\sqrt{\det(L)}}e^{-KS(0)}

Thus what we were missing above when we took out the zero modes, was to divide by a volume factor Vol(G)Vol𝐺{\mbox{Vol}}(G).

This simple 2d example shows two special cases of a more general result [11]. When the gauge symmetry is fully gauge fixed by a saddle point solution (in the above example, X=R0>0𝑋subscript𝑅00X=R_{0}>0), we get

Z𝑍\displaystyle Z =\displaystyle= eS(saddle point)×(one-loop determinants)superscript𝑒𝑆saddle pointone-loop determinants\displaystyle e^{-S(\mbox{saddle point})}\times({\mbox{one-loop determinants}})

If on the other hand the gauge symmetry is not gauge fixed at all by the saddle-point solution (in the above example, X=Y=0𝑋𝑌0X=Y=0), we get

Z𝑍\displaystyle Z =\displaystyle= 1Vol(𝒢)×eS(saddle point)×(one-loop determinants)1Vol𝒢superscript𝑒𝑆saddle pointone-loop determinants\displaystyle\frac{1}{{\mbox{Vol}}({\cal{G}})}\times e^{-S(\mbox{saddle point})}\times({\mbox{one-loop determinants}})

In quantum field theories, the gauge symmetry is an infinite-dimensional local symmetry at each point on the manifold and then we denote such a gauge symmetry as 𝒢𝒢{\cal{G}} which is a G𝐺G-bundle over the manifold. But we can consider quantum theories that are not field theories, and whose gauge symmetry is not a local symmetry but can be any reduntant description of the quantum theory. For gauge groups bigger than SO(2)𝑆𝑂2SO(2), there can also be intermediate cases where the gauge symmetry is only partially gauge fixed. For those cases we have

Z𝑍\displaystyle Z =\displaystyle= 1Vol()×eS(saddle point)×(one-loop determinants)1Volsuperscript𝑒𝑆saddle pointone-loop determinants\displaystyle\frac{1}{{\mbox{Vol}}({\cal{H}})}\times e^{-S(\mbox{saddle point})}\times({\mbox{one-loop determinants}})

where {\cal{H}} is the subgroup of the gauge symmetry 𝒢𝒢{\cal{G}} that is not gauge fixed by the saddle point solution. To illustrate such a case we need a bigger gauge group than U(1)𝑈1U(1) in order to have a proper subgroup. Let us consider an example with SO(3)𝑆𝑂3SO(3) gauge symmetry,

Z𝑍\displaystyle Z =\displaystyle= 1Vol(G)dX2πdY2πdZ2πeKS(R)1Vol𝐺𝑑𝑋2𝜋𝑑𝑌2𝜋𝑑𝑍2𝜋superscript𝑒𝐾𝑆𝑅\displaystyle\frac{1}{{\mbox{Vol}}(G)}\int\frac{dX}{\sqrt{2\pi}}\frac{dY}{\sqrt{2\pi}}\frac{dZ}{\sqrt{2\pi}}e^{-KS(R)}

The saddle point approximation gives

Z𝑍\displaystyle Z =\displaystyle= 1Vol(G)2R02eKS(R0)KS′′(R0)1Vol𝐺2superscriptsubscript𝑅02superscript𝑒𝐾𝑆subscript𝑅0𝐾superscript𝑆′′subscript𝑅0\displaystyle\frac{1}{{\mbox{Vol}}(G)}\frac{2R_{0}^{2}e^{-KS(R_{0})}}{\sqrt{KS^{\prime\prime}(R_{0})}}

The gauge group G=SO(3)𝐺𝑆𝑂3G=SO(3) that acts as

(XΛYΛZΛ)matrixsuperscript𝑋Λsuperscript𝑌Λsuperscript𝑍Λ\displaystyle\begin{pmatrix}X^{\Lambda}\\ Y^{\Lambda}\\ Z^{\Lambda}\end{pmatrix} =\displaystyle= (cosαsinα0sinαcosα0001)(cosβ0sinβ010sinβ0cosβ)(cosγsinγ0sinγcosγ0001)(XYZ)matrix𝛼𝛼0𝛼𝛼0001matrix𝛽0𝛽010𝛽0𝛽matrix𝛾𝛾0𝛾𝛾0001matrix𝑋𝑌𝑍\displaystyle\begin{pmatrix}\cos\alpha&-\sin\alpha&0\\ \sin\alpha&\cos\alpha&0\\ 0&0&1\end{pmatrix}\begin{pmatrix}\cos\beta&0&\sin\beta\\ 0&1&0\\ -\sin\beta&0&\cos\beta\end{pmatrix}\begin{pmatrix}\cos\gamma&-\sin\gamma&0\\ \sin\gamma&\cos\gamma&0\\ 0&0&1\end{pmatrix}\begin{pmatrix}X\\ Y\\ Z\end{pmatrix}

Here we have the SO(3)𝑆𝑂3SO(3) coordinate ranges, α[0,2π]𝛼02𝜋\alpha\in[0,2\pi], β[0,π]𝛽0𝜋\beta\in[0,\pi] and γ[0,2π]𝛾02𝜋\gamma\in[0,2\pi]. If we fix the gauge X=Y=0𝑋𝑌0X=Y=0, there will be a residual gauge symmetry H=SO(2)𝐻𝑆𝑂2H=SO(2) whose rotations are parametrized by the angle γ𝛾\gamma. The rotations of the points at X=Y=0𝑋𝑌0X=Y=0 (the north and the south poles) are given by

(XΛYΛZΛ)matrixsuperscript𝑋Λsuperscript𝑌Λsuperscript𝑍Λ\displaystyle\begin{pmatrix}X^{\Lambda}\\ Y^{\Lambda}\\ Z^{\Lambda}\end{pmatrix} =\displaystyle= (ZcosαsinβZsinαsinβZcosβ)matrix𝑍𝛼𝛽𝑍𝛼𝛽𝑍𝛽\displaystyle\begin{pmatrix}Z\cos\alpha\sin\beta\\ Z\sin\alpha\sin\beta\\ Z\cos\beta\end{pmatrix}

To fix the gauge partially by imposing X=Y=0𝑋𝑌0X=Y=0, we define two gauge fixing functions

G1Λsubscriptsuperscript𝐺Λ1\displaystyle G^{\Lambda}_{1} =\displaystyle= KR0XΛ𝐾subscript𝑅0superscript𝑋Λ\displaystyle KR_{0}X^{\Lambda}
G2Λsubscriptsuperscript𝐺Λ2\displaystyle G^{\Lambda}_{2} =\displaystyle= KR0YΛ𝐾subscript𝑅0superscript𝑌Λ\displaystyle KR_{0}Y^{\Lambda}

We have

1=𝑑G1Λ𝑑G2Λδ(G1Λ)δ(G2Λ)=𝑑α𝑑βJδ(G1Λ)δ(G2Λ)1differential-dsubscriptsuperscript𝐺Λ1differential-dsubscriptsuperscript𝐺Λ2𝛿subscriptsuperscript𝐺Λ1𝛿subscriptsuperscript𝐺Λ2differential-d𝛼differential-d𝛽𝐽𝛿subscriptsuperscript𝐺Λ1𝛿subscriptsuperscript𝐺Λ2\displaystyle 1=\int dG^{\Lambda}_{1}dG^{\Lambda}_{2}\delta(G^{\Lambda}_{1})\delta(G^{\Lambda}_{2})=\int d\alpha d\beta J\delta(G^{\Lambda}_{1})\delta(G^{\Lambda}_{2})

where J𝐽J is the Jacobian

J𝐽\displaystyle J =\displaystyle= |K2R02(αXΛβYΛβXΛαYΛ)|superscript𝐾2superscriptsubscript𝑅02subscript𝛼superscript𝑋Λsubscript𝛽superscript𝑌Λsubscript𝛽superscript𝑋Λsubscript𝛼superscript𝑌Λ\displaystyle|K^{2}R_{0}^{2}\left(\partial_{\alpha}X^{\Lambda}\partial_{\beta}Y^{\Lambda}-\partial_{\beta}X^{\Lambda}\partial_{\alpha}Y^{\Lambda}\right)|

At the points X=Y=0𝑋𝑌0X=Y=0 this becomes

J𝐽\displaystyle J =\displaystyle= K2R04|sinβcosβ|superscript𝐾2superscriptsubscript𝑅04𝛽𝛽\displaystyle K^{2}R_{0}^{4}|\sin\beta\cos\beta|

We then write

dαdβdγJ=K2R04𝒟Λcosβ𝑑𝛼𝑑𝛽𝑑𝛾𝐽superscript𝐾2superscriptsubscript𝑅04𝒟Λ𝛽\displaystyle d\alpha d\beta d\gamma J=K^{2}R_{0}^{4}{\cal{D}}\Lambda\cos\beta

where

𝒟Λ=dαdβdγsinβ𝒟Λ𝑑𝛼𝑑𝛽𝑑𝛾𝛽\displaystyle{\cal{D}}\Lambda=d\alpha d\beta d\gamma\sin\beta

is the Haar measure of SO(3)𝑆𝑂3SO(3). Then we make a gauge rotation of the action and use the gauge invariance, which enables us to isolate an integral over the Haar measure alone, and put α=β=γ=0𝛼𝛽𝛾0\alpha=\beta=\gamma=0 everywhere else. This way we get

detFPsubscript𝐹𝑃\displaystyle\det\triangle_{FP} =\displaystyle= K2R02superscript𝐾2superscriptsubscript𝑅02\displaystyle K^{2}R_{0}^{2}

and then we end up with the result

Z=1Vol(H)detFPdet(L)eKS(R0)𝑍1Vol𝐻subscript𝐹𝑃𝐿superscript𝑒𝐾𝑆subscript𝑅0\displaystyle Z=\frac{1}{{\mbox{Vol}}(H)}\frac{\det\triangle_{FP}}{\sqrt{\det(L)}}e^{-KS(R_{0})}

where

L𝐿\displaystyle L =\displaystyle= K(002πiR0000002πiR002πiR0000002πiR00000000S′′(R0))𝐾matrix002𝜋𝑖subscript𝑅0000002𝜋𝑖subscript𝑅002𝜋𝑖subscript𝑅0000002𝜋𝑖subscript𝑅00000000superscript𝑆′′subscript𝑅0\displaystyle K\begin{pmatrix}0&0&2\pi iR_{0}&0&0\\ 0&0&0&2\pi iR_{0}&0\\ 2\pi iR_{0}&0&0&0&0\\ 0&2\pi iR_{0}&0&0&0\\ 0&0&0&0&S^{\prime\prime}(R_{0})\end{pmatrix}

and

Vol(H)=02π𝑑γVol𝐻superscriptsubscript02𝜋differential-d𝛾\displaystyle{\mbox{Vol}}(H)=\int_{0}^{2\pi}d\gamma

By explicitly computing this expression for Z𝑍Z, we reproduce the result of the saddle-point approximation.

Appendix D Gauge fixing of zero modes

Gauge fixing of fermionic and bosonic zero modes has been analysed in [13]. This method has reappeared more recently in supersymmetric localization [6, 4]. Our topological field theories in 6d and 5d consist of fields with corresponding ghost hierarchy that are all p𝑝p-forms of various degrees, either fermionic or bosonic. Let us assume the gauge group is abelian. Then by Hodge decomposition, any bosonic p𝑝p-form can be decomposed into a coexact, an exact and a harmonic piece,

Apsubscript𝐴𝑝\displaystyle A_{p} =\displaystyle= dαp1+dβp1+γpsuperscript𝑑subscript𝛼𝑝1𝑑subscript𝛽𝑝1subscript𝛾𝑝\displaystyle d^{{\dagger}}\alpha_{p-1}+d\beta_{p-1}+\gamma_{p}

D.1 Bosonic zero mode gauge fixing

If the action has a gauge symmetry δAp=dΛp1𝛿subscript𝐴𝑝𝑑subscriptΛ𝑝1\delta A_{p}=d\Lambda_{p-1}, then βp1subscript𝛽𝑝1\beta_{p-1} is projected out by gauge fixing. There can also be zero modes, which we will treat in a similar way as the above gauge symmetries. A zero mode for Apsubscript𝐴𝑝A_{p}, means that the action is invariant under δAp=Λp𝛿subscript𝐴𝑝subscriptΛ𝑝\delta A_{p}=\Lambda_{p} where ΛpsubscriptΛ𝑝\Lambda_{p} is harmonic. We treat this as a gauge symmetry that we gauge fix by adding the Lagrange multiplier term ϵi(ωpi,A)subscriptitalic-ϵ𝑖subscriptsuperscript𝜔𝑖𝑝𝐴\epsilon_{i}(\omega^{i}_{p},A) to the action. Here ϵisubscriptitalic-ϵ𝑖\epsilon_{i} are bosonic constant Lagrange multipliers, ωpisubscriptsuperscript𝜔𝑖𝑝\omega^{i}_{p} is some metric-independent choice of basis for the space of harmonic p𝑝p-forms. Integrating over ϵisubscriptitalic-ϵ𝑖\epsilon_{i} imposes the delta function constraint (ωpi,A)=0subscriptsuperscript𝜔𝑖𝑝𝐴0(\omega^{i}_{p},A)=0, which means the harmonic piece γpsubscript𝛾𝑝\gamma_{p} is projected out in a BRST invariant manner. Here the BRST variations are

δAp𝛿subscript𝐴𝑝\displaystyle\delta A_{p} =\displaystyle= ωpicisubscriptsuperscript𝜔𝑖𝑝subscript𝑐𝑖\displaystyle\omega^{i}_{p}c_{i}
δci𝛿subscript𝑐𝑖\displaystyle\delta c_{i} =\displaystyle= 00\displaystyle 0
δc¯i𝛿subscript¯𝑐𝑖\displaystyle\delta\bar{c}_{i} =\displaystyle= ϵisubscriptitalic-ϵ𝑖\displaystyle\epsilon_{i}
δϵi𝛿subscriptitalic-ϵ𝑖\displaystyle\delta\epsilon_{i} =\displaystyle= 00\displaystyle 0

where δ𝛿\delta changes the Grassmann properties of the fields, cisubscript𝑐𝑖c_{i}, c¯isubscript¯𝑐𝑖\bar{c}_{i} and ϵisubscriptitalic-ϵ𝑖\epsilon_{i} are all constants. The full BRST exact gauge fixing term is

δ(c¯iωpi,Ap)=ϵi(ωpi,Ap)c¯icj(ωpi,ωpj)𝛿subscript¯𝑐𝑖subscriptsuperscript𝜔𝑖𝑝subscript𝐴𝑝subscriptitalic-ϵ𝑖subscriptsuperscript𝜔𝑖𝑝subscript𝐴𝑝subscript¯𝑐𝑖subscript𝑐𝑗subscriptsuperscript𝜔𝑖𝑝subscriptsuperscript𝜔𝑗𝑝\displaystyle\delta(\bar{c}_{i}\omega^{i}_{p},A_{p})=\epsilon_{i}(\omega^{i}_{p},A_{p})-\bar{c}_{i}c_{j}(\omega^{i}_{p},\omega^{j}_{p})

We then first consider the path integral over the bosonic zero modes

[𝒩BdAi][𝒩Bdϵi]eϵiAj(ωpi,ωpj)delimited-[]subscript𝒩𝐵𝑑subscript𝐴𝑖delimited-[]subscript𝒩𝐵𝑑subscriptitalic-ϵ𝑖superscript𝑒subscriptitalic-ϵ𝑖subscript𝐴𝑗superscriptsubscript𝜔𝑝𝑖superscriptsubscript𝜔𝑝𝑗\displaystyle\int[{\cal{N}}_{B}dA_{i}][{\cal{N}}_{B}d\epsilon_{i}]e^{\epsilon_{i}A_{j}(\omega_{p}^{i},\omega_{p}^{j})} =\displaystyle= 𝒩B2[dAi]2πδ(Aj(ωpi,ωpj))superscriptsubscript𝒩𝐵2delimited-[]𝑑subscript𝐴𝑖2𝜋𝛿subscript𝐴𝑗superscriptsubscript𝜔𝑝𝑖superscriptsubscript𝜔𝑝𝑗\displaystyle{\cal{N}}_{B}^{2}\int[dA_{i}]2\pi\delta(A_{j}(\omega_{p}^{i},\omega_{p}^{j}))
=\displaystyle= 2πvp𝒩B2[dAi]δ(Ai)2𝜋subscript𝑣𝑝superscriptsubscript𝒩𝐵2delimited-[]𝑑superscript𝐴𝑖𝛿superscript𝐴𝑖\displaystyle\frac{2\pi}{v_{p}}{\cal{N}}_{B}^{2}\int[dA^{i}]\delta(A^{i})
=\displaystyle= 2πvp𝒩B22𝜋subscript𝑣𝑝superscriptsubscript𝒩𝐵2\displaystyle\frac{2\pi}{v_{p}}{\cal{N}}_{B}^{2}

where vp=det(ωpi,ωpj)subscript𝑣𝑝superscriptsubscript𝜔𝑝𝑖superscriptsubscript𝜔𝑝𝑗v_{p}=\det(\omega_{p}^{i},\omega_{p}^{j}) is the Jacobian that is produced as we change variables from Aisubscript𝐴𝑖A_{i} to Ai=(ωpi,ωpj)Ajsuperscript𝐴𝑖subscriptsuperscript𝜔𝑖𝑝subscriptsuperscript𝜔𝑗𝑝subscript𝐴𝑗A^{i}=(\omega^{i}_{p},\omega^{j}_{p})A_{j} in the measure. Next we consider the path integral over the fermionic zero modes

[𝒩Fdci][𝒩Fdc¯i]ec¯icj(ωpi,ωpj)delimited-[]subscript𝒩𝐹𝑑subscript𝑐𝑖delimited-[]subscript𝒩𝐹𝑑subscript¯𝑐𝑖superscript𝑒subscript¯𝑐𝑖subscript𝑐𝑗subscriptsuperscript𝜔𝑖𝑝subscriptsuperscript𝜔𝑗𝑝\displaystyle\int[{\cal{N}}_{F}dc_{i}][{\cal{N}}_{F}d\overline{c}_{i}]e^{\bar{c}_{i}c_{j}(\omega^{i}_{p},\omega^{j}_{p})} =\displaystyle= vp𝒩F2subscript𝑣𝑝superscriptsubscript𝒩𝐹2\displaystyle v_{p}{\cal{N}}_{F}^{2}

Multiplying together, we get

2π(𝒩B𝒩F)22𝜋superscriptsubscript𝒩𝐵subscript𝒩𝐹2\displaystyle 2\pi({\cal{N}}_{B}{\cal{N}}_{F})^{2}

D.2 Fermionic zero mode gauge fixing

If instead the p𝑝p-form is a fermionic field ψpsubscript𝜓𝑝\psi_{p} with the symmetry δψp=λp𝛿subscript𝜓𝑝subscript𝜆𝑝\delta\psi_{p}=\lambda_{p} where λpsubscript𝜆𝑝\lambda_{p} is a fermionic harmonic p𝑝p-form, then we add the Lagrange multiplier term ϵi(ωpi,ψp)subscriptitalic-ϵ𝑖subscriptsuperscript𝜔𝑖𝑝subscript𝜓𝑝\epsilon_{i}(\omega^{i}_{p},\psi_{p}) to the action where now ϵisubscriptitalic-ϵ𝑖\epsilon_{i} are fermionic constant parameters. BRST variations are

δψp𝛿subscript𝜓𝑝\displaystyle\delta\psi_{p} =\displaystyle= aiωpisubscript𝑎𝑖superscriptsubscript𝜔𝑝𝑖\displaystyle a_{i}\omega_{p}^{i}
δai𝛿subscript𝑎𝑖\displaystyle\delta a_{i} =\displaystyle= 00\displaystyle 0
δa¯i𝛿subscript¯𝑎𝑖\displaystyle\delta\bar{a}_{i} =\displaystyle= ϵisubscriptitalic-ϵ𝑖\displaystyle\epsilon_{i}
δϵi𝛿subscriptitalic-ϵ𝑖\displaystyle\delta\epsilon_{i} =\displaystyle= 00\displaystyle 0

where ϵisubscriptitalic-ϵ𝑖\epsilon_{i} are fermionic zero modes, ai,a¯isubscript𝑎𝑖subscript¯𝑎𝑖a_{i},\overline{a}_{i} are bosonic zero modes. We add the BRST-exact term

δ(𝒩1(a¯iωpi,ψ)+𝒩2(aiωpi,ψ))𝛿subscript𝒩1subscript¯𝑎𝑖superscriptsubscript𝜔𝑝𝑖𝜓subscript𝒩2subscript𝑎𝑖superscriptsubscript𝜔𝑝𝑖𝜓\displaystyle\delta\big{(}{\cal{N}}_{1}(\overline{a}_{i}\omega_{p}^{i},\psi)+{\cal{N}}_{2}(a_{i}\omega_{p}^{i},\psi)\big{)} =\displaystyle= 𝒩1ϵi(ωpi,ψ)+𝒩1a¯iaj(ωpi,ωpj)+𝒩2aiaj(ωpi,ωpj)subscript𝒩1subscriptitalic-ϵ𝑖superscriptsubscript𝜔𝑝𝑖𝜓subscript𝒩1subscript¯𝑎𝑖subscript𝑎𝑗superscriptsubscript𝜔𝑝𝑖superscriptsubscript𝜔𝑝𝑗subscript𝒩2subscript𝑎𝑖subscript𝑎𝑗superscriptsubscript𝜔𝑝𝑖superscriptsubscript𝜔𝑝𝑗\displaystyle{\cal{N}}_{1}\epsilon_{i}(\omega_{p}^{i},\psi)+{\cal{N}}_{1}\overline{a}_{i}a_{j}(\omega_{p}^{i},\omega_{p}^{j})+{\cal{N}}_{2}a_{i}a_{j}(\omega_{p}^{i},\omega_{p}^{j})
=\displaystyle= (𝒩1ϵiψj+𝒩1a¯iaj+𝒩2aiaj)(ωpi,ωpj)subscript𝒩1subscriptitalic-ϵ𝑖subscript𝜓𝑗subscript𝒩1subscript¯𝑎𝑖subscript𝑎𝑗subscript𝒩2subscript𝑎𝑖subscript𝑎𝑗subscriptsuperscript𝜔𝑖𝑝subscriptsuperscript𝜔𝑗𝑝\displaystyle\big{(}{\cal{N}}_{1}\epsilon_{i}\psi_{j}+{\cal{N}}_{1}\overline{a}_{i}a_{j}+{\cal{N}}_{2}a_{i}a_{j}\big{)}(\omega^{i}_{p},\omega^{j}_{p})

The path integral over the fermionic zero modes is

Z0subscript𝑍0\displaystyle Z_{0} =\displaystyle= [𝒩Fdψi][𝒩Fdϵi]exp[(𝒩1ϵiψj+𝒩1a¯iaj+𝒩2aiaj)(ωpi,ωpj)]delimited-[]subscript𝒩𝐹𝑑subscript𝜓𝑖delimited-[]subscript𝒩𝐹𝑑subscriptitalic-ϵ𝑖subscript𝒩1subscriptitalic-ϵ𝑖subscript𝜓𝑗subscript𝒩1subscript¯𝑎𝑖subscript𝑎𝑗subscript𝒩2subscript𝑎𝑖subscript𝑎𝑗subscriptsuperscript𝜔𝑖𝑝subscriptsuperscript𝜔𝑗𝑝\displaystyle\int[{\cal{N}}_{F}d\psi_{i}][{\cal{N}}_{F}d\epsilon_{i}]\exp\left[-\big{(}{\cal{N}}_{1}\epsilon_{i}\psi_{j}+{\cal{N}}_{1}\overline{a}_{i}a_{j}+{\cal{N}}_{2}a_{i}a_{j}\big{)}(\omega^{i}_{p},\omega^{j}_{p})\right]
=\displaystyle= (𝒩F)2𝒩1vpexp[(𝒩1a¯iaj+𝒩2aiaj)(ωpi,ωpj)]superscriptsubscript𝒩𝐹2subscript𝒩1subscript𝑣𝑝subscript𝒩1subscript¯𝑎𝑖subscript𝑎𝑗subscript𝒩2subscript𝑎𝑖subscript𝑎𝑗subscriptsuperscript𝜔𝑖𝑝subscriptsuperscript𝜔𝑗𝑝\displaystyle({\cal{N}}_{F})^{2}{\cal{N}}_{1}v_{p}\exp\left[-\big{(}{\cal{N}}_{1}\overline{a}_{i}a_{j}+{\cal{N}}_{2}a_{i}a_{j}\big{)}(\omega^{i}_{p},\omega^{j}_{p})\right]

We complete the square,

(ωpi,ωpj)(𝒩1a¯iaj+𝒩2aiaj)subscriptsuperscript𝜔𝑖𝑝subscriptsuperscript𝜔𝑗𝑝subscript𝒩1subscript¯𝑎𝑖subscript𝑎𝑗subscript𝒩2subscript𝑎𝑖subscript𝑎𝑗\displaystyle(\omega^{i}_{p},\omega^{j}_{p})\left({\cal{N}}_{1}\overline{a}_{i}a_{j}+{\cal{N}}_{2}a_{i}a_{j}\right) =\displaystyle= 𝒩2(ωpi,ωpj)(ai+𝒩12𝒩2a¯i)(aj+𝒩12𝒩2a¯j)(𝒩1)24𝒩2(ωpi,ωpj)a¯ia¯jsubscript𝒩2subscriptsuperscript𝜔𝑖𝑝subscriptsuperscript𝜔𝑗𝑝subscript𝑎𝑖subscript𝒩12subscript𝒩2subscript¯𝑎𝑖subscript𝑎𝑗subscript𝒩12subscript𝒩2subscript¯𝑎𝑗superscriptsubscript𝒩124subscript𝒩2subscriptsuperscript𝜔𝑖𝑝subscriptsuperscript𝜔𝑗𝑝subscript¯𝑎𝑖subscript¯𝑎𝑗\displaystyle{\cal{N}}_{2}(\omega^{i}_{p},\omega^{j}_{p})\left(a_{i}+\frac{{\cal{N}}_{1}}{2{\cal{N}}_{2}}\overline{a}_{i}\right)\left(a_{j}+\frac{{\cal{N}}_{1}}{2{\cal{N}}_{2}}\overline{a}_{j}\right)-\frac{({\cal{N}}_{1})^{2}}{4{\cal{N}}_{2}}(\omega^{i}_{p},\omega^{j}_{p})\overline{a}_{i}\overline{a}_{j}

The Gaussian integral is convergent for 𝒩2>0subscript𝒩20{\cal{N}}_{2}>0 and 𝒩1subscript𝒩1{\cal{N}}_{1} purely imaginary. At such values we can compute the Gaussian integrals over the bosonic zero modes

[𝒩Bdai][𝒩Bda¯i]exp(𝒩2(ai+𝒩12𝒩2a¯i)2+(𝒩1)24𝒩2(a¯i)2)delimited-[]subscript𝒩𝐵𝑑subscript𝑎𝑖delimited-[]subscript𝒩𝐵𝑑subscript¯𝑎𝑖subscript𝒩2superscriptsubscript𝑎𝑖subscript𝒩12subscript𝒩2subscript¯𝑎𝑖2superscriptsubscript𝒩124subscript𝒩2superscriptsubscript¯𝑎𝑖2\displaystyle\int[{\cal{N}}_{B}da_{i}][{\cal{N}}_{B}d\overline{a}_{i}]\exp\left(-{\cal{N}}_{2}\left(a_{i}+\frac{{\cal{N}}_{1}}{2{\cal{N}}_{2}}\overline{a}_{i}\right)^{2}+\frac{({\cal{N}}_{1})^{2}}{4{\cal{N}}_{2}}(\overline{a}_{i})^{2}\right)

and get the result

Z0=(𝒩B𝒩F)2𝒩1vpπ𝒩2vpπ(𝒩1)24𝒩2vp=2π(𝒩B𝒩F)2subscript𝑍0superscriptsubscript𝒩𝐵subscript𝒩𝐹2subscript𝒩1subscript𝑣𝑝𝜋subscript𝒩2subscript𝑣𝑝𝜋superscriptsubscript𝒩124subscript𝒩2subscript𝑣𝑝2𝜋superscriptsubscript𝒩𝐵subscript𝒩𝐹2\displaystyle Z_{0}=({\cal{N}}_{B}{\cal{N}}_{F})^{2}{\cal{N}}_{1}v_{p}\sqrt{\frac{\pi}{{\cal{N}}_{2}v_{p}}}\sqrt{\frac{\pi}{\frac{({\cal{N}}_{1})^{2}}{4{\cal{N}}_{2}}v_{p}}}=2\pi({\cal{N}}_{B}{\cal{N}}_{F})^{2}

We see that all the dependence on 𝒩1subscript𝒩1{\cal{N}}_{1} and 𝒩2subscript𝒩2{\cal{N}}_{2} cancels out. This was known by general considerations since the added term was BRST-exact, but it is nevertheless nice to see how this happens by an explicit computation. At other values of 𝒩1subscript𝒩1{\cal{N}}_{1} and 𝒩2subscript𝒩2{\cal{N}}_{2} we define the path integral by analytic continuation. Since it is just a constant, the analytic continuation of the path integral is trivial – it will remain to be equal to this constant value for all values on 𝒩1subscript𝒩1{\cal{N}}_{1} and 𝒩2subscript𝒩2{\cal{N}}_{2}.

D.3 Fermionic zero mode gauge fixing, once again

As was noted in [13], this method does not work for all the p𝑝p-forms in a ghost hierarchy. To quote [13]: ‘This works well for the ghosts that are present on the right-hand ledge of the ghost-triangle.’ To illustrate what is meant by this, let us consider as an example Maxwell theory with the nonharmonic BRST variations

δA𝛿𝐴\displaystyle\delta A =\displaystyle= dc𝑑𝑐\displaystyle dc (D.1)
δc¯𝛿¯𝑐\displaystyle\delta\bar{c} =\displaystyle= iB𝑖𝐵\displaystyle iB (D.2)
δB𝛿𝐵\displaystyle\delta B =\displaystyle= 00\displaystyle 0 (D.3)
δc𝛿𝑐\displaystyle\delta c =\displaystyle= 00\displaystyle 0 (D.4)

There are two ghosts c𝑐c and c¯¯𝑐\bar{c}, but only the ghost c𝑐c is on the right-ledge of the ghost-triangle. Let us assume these have zero-form harmonics with corresponding BRST variations. For the c𝑐c ghost, these will be

δc𝛿𝑐\displaystyle\delta c =\displaystyle= aiω0isubscript𝑎𝑖superscriptsubscript𝜔0𝑖\displaystyle a_{i}\omega_{0}^{i}
δai𝛿subscript𝑎𝑖\displaystyle\delta a_{i} =\displaystyle= 00\displaystyle 0
δa¯i𝛿subscript¯𝑎𝑖\displaystyle\delta\bar{a}_{i} =\displaystyle= ϵisubscriptitalic-ϵ𝑖\displaystyle\epsilon_{i}
δϵi𝛿subscriptitalic-ϵ𝑖\displaystyle\delta\epsilon_{i} =\displaystyle= 00\displaystyle 0

which remain nilpotent also when combined with (D.4). But for the c¯¯𝑐\bar{c} ghost we already have BRST transformations from the above

δc¯𝛿¯𝑐\displaystyle\delta\bar{c} =\displaystyle= iB𝑖𝐵\displaystyle iB
δB𝛿𝐵\displaystyle\delta B =\displaystyle= 00\displaystyle 0

which can be extended to include harmonic parts as well. We then enlarge this by adding (constant) ghosts σ¯¯𝜎\bar{\sigma} and τ𝜏\tau whose BRST variations are

δσ¯i𝛿subscript¯𝜎𝑖\displaystyle\delta\bar{\sigma}_{i} =\displaystyle= iτi𝑖subscript𝜏𝑖\displaystyle i\tau_{i}
δτi𝛿subscript𝜏𝑖\displaystyle\delta\tau_{i} =\displaystyle= 00\displaystyle 0

Then we add the BRST exact term

δ(σ¯iω0i,c¯)=i(τiω0i,c¯)+i(σ¯iω0i,B)𝛿subscript¯𝜎𝑖subscriptsuperscript𝜔𝑖0¯𝑐𝑖subscript𝜏𝑖subscriptsuperscript𝜔𝑖0¯𝑐𝑖subscript¯𝜎𝑖subscriptsuperscript𝜔𝑖0𝐵\displaystyle\delta(\bar{\sigma}_{i}\omega^{i}_{0},\bar{c})=i(\tau_{i}\omega^{i}_{0},\bar{c})+i(\bar{\sigma}_{i}\omega^{i}_{0},B)

When we integrate over the fermionic zero modes, we get

[𝒩Fdτi][𝒩Fdc¯i]eiτic¯j(ω0i,ω0j)delimited-[]subscript𝒩𝐹𝑑subscript𝜏𝑖delimited-[]subscript𝒩𝐹𝑑subscript¯𝑐𝑖superscript𝑒𝑖subscript𝜏𝑖subscript¯𝑐𝑗superscriptsubscript𝜔0𝑖superscriptsubscript𝜔0𝑗\displaystyle\int[{\cal{N}}_{F}d\tau_{i}][{\cal{N}}_{F}d\overline{c}_{i}]e^{i\tau_{i}\bar{c}_{j}(\omega_{0}^{i},\omega_{0}^{j})} =\displaystyle= ivp(𝒩F)2𝑖subscript𝑣𝑝superscriptsubscript𝒩𝐹2\displaystyle iv_{p}({\cal{N}}_{F})^{2}

For the bosons, we get

[𝒩Bdσi][𝒩BdBi]eiσ¯iBj(ω0i,ω0j)delimited-[]subscript𝒩𝐵𝑑subscript𝜎𝑖delimited-[]subscript𝒩𝐵𝑑subscript𝐵𝑖superscript𝑒𝑖subscript¯𝜎𝑖subscript𝐵𝑗superscriptsubscript𝜔0𝑖superscriptsubscript𝜔0𝑗\displaystyle\int[{\cal{N}}_{B}d\sigma_{i}][{\cal{N}}_{B}dB_{i}]e^{i\overline{\sigma}_{i}B_{j}(\omega_{0}^{i},\omega_{0}^{j})} =\displaystyle= (𝒩B)2[dBi]2πδ(Bj(ω0i,ω0j))superscriptsubscript𝒩𝐵2delimited-[]𝑑subscript𝐵𝑖2𝜋𝛿subscript𝐵𝑗superscriptsubscript𝜔0𝑖superscriptsubscript𝜔0𝑗\displaystyle({\cal{N}}_{B})^{2}\int[dB_{i}]2\pi\delta(B_{j}(\omega_{0}^{i},\omega_{0}^{j}))
=\displaystyle= 2π(𝒩B)2vp2𝜋superscriptsubscript𝒩𝐵2subscript𝑣𝑝\displaystyle\frac{2\pi({\cal{N}}_{B})^{2}}{v_{p}}

Multiplying together, we get

2π(𝒩F𝒩B)22𝜋superscriptsubscript𝒩𝐹subscript𝒩𝐵2\displaystyle 2\pi({\cal{N}}_{F}{\cal{N}}_{B})^{2}

If we choose the path integral measure for the zero modes such that 𝒩F𝒩B=12πsubscript𝒩𝐹subscript𝒩𝐵12𝜋{\cal{N}}_{F}{\cal{N}}_{B}=\frac{1}{\sqrt{2\pi}}, then we can summarize our result as follows: removing any set of harmonic p𝑝p-form zero modes from the path integral in a BRST invariant way, always produces the same factor 2π(𝒩F𝒩B)2=12𝜋superscriptsubscript𝒩𝐹subscript𝒩𝐵212\pi({\cal{N}}_{F}{\cal{N}}_{B})^{2}=1 no matter the zero mode is bosonic or fermionic, or on the right-ledge of the ghost-triangle or not. All sets of harmonic zero modes produce the same factor.

Appendix E A review of 3d Chern-Simons perturbation theory

Here we review what we will need from [1, 2, 8]. The starting point is the Chern-Simons action

S(A)𝑆𝐴\displaystyle S(A) =\displaystyle= k4πtr(AdA2i3A3)𝑘4𝜋tr𝐴𝑑𝐴2𝑖3superscript𝐴3\displaystyle\frac{k}{4\pi}\int{\mbox{tr}}\left(A\wedge dA-\frac{2i}{3}A^{3}\right)

If we define the covariant derivative as

DmAsuperscriptsubscript𝐷𝑚𝐴\displaystyle D_{m}^{A} =\displaystyle= mi[Am,]subscript𝑚𝑖subscript𝐴𝑚\displaystyle\nabla_{m}-i[A_{m},\cdot]

then a gauge transformation associated with the group element g𝑔g will act as

Amsubscript𝐴𝑚\displaystyle A_{m} \displaystyle\rightarrow Amgsuperscriptsubscript𝐴𝑚𝑔\displaystyle A_{m}^{g}
Amgsuperscriptsubscript𝐴𝑚𝑔\displaystyle A_{m}^{g} =\displaystyle= ig1mg+g1Amg𝑖superscript𝑔1subscript𝑚𝑔superscript𝑔1subscript𝐴𝑚𝑔\displaystyle ig^{-1}\nabla_{m}g+g^{-1}A_{m}g

This can also be expressed as

DmAgsuperscriptsubscript𝐷𝑚superscript𝐴𝑔\displaystyle D_{m}^{A^{g}} =\displaystyle= g1DmAgsuperscript𝑔1superscriptsubscript𝐷𝑚𝐴𝑔\displaystyle g^{-1}D_{m}^{A}g

We have BRST variations

δAm𝛿subscript𝐴𝑚\displaystyle\delta A_{m} =\displaystyle= Dmcsubscript𝐷𝑚𝑐\displaystyle D_{m}c
δc𝛿𝑐\displaystyle\delta c =\displaystyle= i2{c,c}𝑖2𝑐𝑐\displaystyle\frac{i}{2}\{c,c\}
δB𝛿𝐵\displaystyle\delta B =\displaystyle= 00\displaystyle 0
δc¯𝛿¯𝑐\displaystyle\delta\overline{c} =\displaystyle= iB𝑖𝐵\displaystyle iB

The partition function can be computed perturbatively in 1/k1𝑘1/k by expanding the action to quadratic order around the saddle points.

One expands the gauge potential around a flat connection A()superscript𝐴A^{(\ell)},

Amsubscript𝐴𝑚\displaystyle A_{m} =\displaystyle= Am()+amsubscriptsuperscript𝐴𝑚subscript𝑎𝑚\displaystyle A^{(\ell)}_{m}+a_{m}

Since we are not interested in gauge transforming the flat connection to zero (if we do that, then we change the boundary conditions of the fields), it is natural to impose the following gauge transformation rules for these new fields,

A()mgsubscriptsuperscriptsuperscript𝐴𝑔𝑚\displaystyle{A^{(\ell)}}^{g}_{m} =\displaystyle= g1Am()gsuperscript𝑔1subscriptsuperscript𝐴𝑚𝑔\displaystyle g^{-1}A^{(\ell)}_{m}g
amgsubscriptsuperscript𝑎𝑔𝑚\displaystyle a^{g}_{m} =\displaystyle= ig1mg+g1amg𝑖superscript𝑔1subscript𝑚𝑔superscript𝑔1subscript𝑎𝑚𝑔\displaystyle ig^{-1}\nabla_{m}g+g^{-1}a_{m}g

meaning that we can only rotate the flat connection (in particular we can diagonalize it), but not gauge transform it to zero. On the other hand, the fluctuation field is now a gauge potential that we need to gauge fix. We define a derivative

Dm():=mi[A(),]assignsubscriptsuperscript𝐷𝑚subscript𝑚𝑖superscript𝐴\displaystyle D^{(\ell)}_{m}:=\nabla_{m}-i[A^{(\ell)},\cdot]

and consider the following nilpotent BRST variations

δam𝛿subscript𝑎𝑚\displaystyle\delta a_{m} =\displaystyle= Dm()csuperscriptsubscript𝐷𝑚𝑐\displaystyle D_{m}^{(\ell)}c
δAm()𝛿subscriptsuperscript𝐴𝑚\displaystyle\delta A^{(\ell)}_{m} =\displaystyle= 00\displaystyle 0
δc𝛿𝑐\displaystyle\delta c =\displaystyle= 00\displaystyle 0
δB𝛿𝐵\displaystyle\delta B =\displaystyle= 00\displaystyle 0
δc¯𝛿¯𝑐\displaystyle\delta\overline{c} =\displaystyle= iB𝑖𝐵\displaystyle iB

We add the gauge fixing term

Sgauge(a)subscript𝑆𝑔𝑎𝑢𝑔𝑒𝑎\displaystyle S_{gauge}(a) =\displaystyle= iδd3xgtr(c¯D()mam)=d3xgtr(BD()mam+ic¯D()mDm()c)𝑖𝛿superscript𝑑3𝑥𝑔tr¯𝑐superscript𝐷𝑚subscript𝑎𝑚superscript𝑑3𝑥𝑔tr𝐵superscript𝐷𝑚subscript𝑎𝑚𝑖¯𝑐superscript𝐷𝑚subscriptsuperscript𝐷𝑚𝑐\displaystyle-i\delta\int d^{3}x\sqrt{g}{\mbox{tr}}\left(\overline{c}D^{(\ell)m}a_{m}\right)=\int d^{3}x\sqrt{g}{\mbox{tr}}\left(BD^{(\ell)m}a_{m}+i\overline{c}D^{(\ell)m}D^{(\ell)}_{m}c\right)

which we will write as

Sgauge(a)subscript𝑆𝑔𝑎𝑢𝑔𝑒𝑎\displaystyle S_{gauge}(a) =\displaystyle= (B,D()a)+i(c¯,0()c)𝐵superscript𝐷𝑎𝑖¯𝑐subscriptsuperscript0𝑐\displaystyle(B,D^{(\ell){\dagger}}a)+i(\overline{c},\triangle^{(\ell)}_{0}c)

We now see that we could also have used the original BRST variations and the full covariant derivative Dmsubscript𝐷𝑚D_{m}. Then we would get the same gauge fixing action Sgaugesubscript𝑆𝑔𝑎𝑢𝑔𝑒S_{gauge} with higher order correction terms, which would play no role for the 1-loop computation.

By multiplying Sgaugesubscript𝑆𝑔𝑎𝑢𝑔𝑒S_{gauge} by an overall constant k2π𝑘2𝜋\frac{k}{2\pi}, the full BRST gauge fixed action becomes of the form

S(a,B,c,c¯)𝑆𝑎𝐵𝑐¯𝑐\displaystyle S(a,B,c,\overline{c}) =\displaystyle= S(A())+k4π(a,Da)+k2π[(B,Da)+i(c¯,0c)]\displaystyle S(A^{(\ell)})+\frac{k}{4\pi}(a,*Da)+\frac{k}{2\pi}\left[(B,D^{{\dagger}}a)+i(\overline{c},\triangle_{0}c)\right]

We can write part of this action as

k4π((aB),(DDD0)(aB))𝑘4𝜋matrix𝑎𝐵matrixabsent𝐷𝐷superscript𝐷0matrix𝑎𝐵\displaystyle\frac{k}{4\pi}\left(\begin{pmatrix}a&B\end{pmatrix},\begin{pmatrix}*D&D\\ D^{{\dagger}}&0\end{pmatrix}\begin{pmatrix}a\\ B\end{pmatrix}\right)

The matrix operator that enters in this expression has the square

(DDD0)(DDD0)matrixabsent𝐷𝐷superscript𝐷0matrixabsent𝐷𝐷superscript𝐷0\displaystyle\begin{pmatrix}*D&D\\ D^{{\dagger}}&0\end{pmatrix}\begin{pmatrix}*D&D\\ D^{{\dagger}}&0\end{pmatrix} =\displaystyle= (1000)matrixsubscript100subscript0\displaystyle\begin{pmatrix}\triangle_{1}&0\\ 0&\triangle_{0}\end{pmatrix} (E.1)

One may also notice that the operator we just squared, is nothing but Lsubscript𝐿L_{-}, which is defined from L=D+DL=*D+D* by restriction to odd forms. If f1subscript𝑓1f_{1} and f3subscript𝑓3f_{3} denote a one-form and a three-form, with coefficients a1subscript𝑎1a_{1} and a3subscript𝑎3a_{3}, then we find that

L(a1f1+a3f3)subscript𝐿subscript𝑎1subscript𝑓1subscript𝑎3subscript𝑓3\displaystyle L_{-}(a_{1}f_{1}+a_{3}f_{3}) =\displaystyle= (a1Df1+a3Df3)+a1Df1subscript𝑎1𝐷subscript𝑓1subscript𝑎3𝐷subscript𝑓3subscript𝑎1𝐷subscript𝑓1\displaystyle\left(a_{1}*Df_{1}+a_{3}D*f_{3}\right)+a_{1}D*f_{1}

If we then write f0=f3f_{0}=*f_{3}, then we find that

L(f1f0)subscript𝐿matrixsubscript𝑓1subscript𝑓0\displaystyle L_{-}\begin{pmatrix}f_{1}\\ f_{0}\end{pmatrix} =\displaystyle= (DDD0)(f1f0)matrixabsent𝐷𝐷superscript𝐷0matrixsubscript𝑓1subscript𝑓0\displaystyle\begin{pmatrix}*D&D\\ D^{{\dagger}}&0\end{pmatrix}\begin{pmatrix}f_{1}\\ f_{0}\end{pmatrix}

The contribution from the flat connection A()superscript𝐴A^{(\ell)} to the partition function becomes

expS(A())det0det(L)𝑆superscript𝐴subscript0subscript𝐿\displaystyle\exp S(A^{(\ell)})\frac{\det\triangle_{0}}{\sqrt{\det(L_{-})}}

Moreover,

1detL1subscript𝐿\displaystyle\frac{1}{\sqrt{\det L_{-}}} =\displaystyle= 1|detL|expiπ2η(A())1subscript𝐿𝑖𝜋2𝜂superscript𝐴\displaystyle\frac{1}{\sqrt{|\det L_{-}|}}\exp\frac{i\pi}{2}\eta(A^{(\ell)})

where, from the APS index theorem,

12η(A())12𝜂superscript𝐴\displaystyle\frac{1}{2}\eta(A^{(\ell)}) =\displaystyle= 12η(0)+c22πS(A())12𝜂0subscript𝑐22𝜋𝑆superscript𝐴\displaystyle\frac{1}{2}\eta(0)+\frac{c_{2}}{2\pi}S(A^{(\ell)})

Thus this phase factor can be absorbed by shifting

kK:=k+c2/2𝑘𝐾assign𝑘subscript𝑐22\displaystyle k\rightarrow K:=k+c_{2}/2

Let us return to the absolute value of the partition function. From (E.1) together with the ghost contribution, we get

det0|det(L)|=(det0)34(det1)14subscript0subscript𝐿superscriptsubscript034superscriptsubscript114\displaystyle\frac{\det\triangle_{0}}{\sqrt{|\det(L_{-})|}}=\left(\det\triangle_{0}\right)^{\frac{3}{4}}\left(\det\triangle_{1}\right)^{-\frac{1}{4}}

If we take away the zero modes, this is the oscillator mode contribution to the square root of the RS torsion.

E.1 The dependence on the Chern-Simons level

To derive the K𝐾K-dependence, all we need to do, is to extract the K𝐾K-dependence from the kinetic term inside the Chern-Simons term. The path integral gives the factor

1det(Kd)121superscript𝐾𝑑12\displaystyle\frac{1}{\det\left(K*d\right)^{\frac{1}{2}}} =\displaystyle= 1K12ζd(0)1det(d)12\displaystyle\frac{1}{K^{\frac{1}{2}\zeta_{*d}(0)}}\frac{1}{\det\left(*d\right)^{\frac{1}{2}}}
ζd(0)subscript𝜁absent𝑑0\displaystyle\zeta_{*d}(0) =\displaystyle= b0b1subscript𝑏0subscript𝑏1\displaystyle b_{0}-b_{1}

If we assume that b1=0subscript𝑏10b_{1}=0, there will be no bosonic zero modes of the operator d𝑑d and the zero mode problem can be avoided. And then this gives the correct K𝐾K-dependence. More specifically, bqsubscript𝑏𝑞b_{q} is the dimension of Hq(M3)subscript𝐻𝑞subscript𝑀3H_{q}(M_{3}) times the dimension of the unbroken gauge group in the background of the flat connection A()superscript𝐴A^{(\ell)}.

Let us finally review the computation of the perturbative partition function for G=SU(2)𝐺𝑆𝑈2G=SU(2) gauge group on lens space S3/psuperscript𝑆3subscript𝑝S^{3}/\mathbb{Z}_{p}. There are flat connections A()superscript𝐴A^{(\ell)} for =0,1,2,,p1012𝑝1\ell=0,1,2,...,p-1. We shall divide by the isotropy group of unbroken gauge symmetries when we turn on the flat connection. When =00\ell=0 the isotropy group is HA(0)=SU(2)subscript𝐻superscript𝐴0𝑆𝑈2H_{A^{(0)}}=SU(2) as no gauge symmetry is broken. When >00\ell>0 the isotropy group is HA()=U(1)subscript𝐻superscript𝐴𝑈1H_{A^{(\ell)}}=U(1).

Since the classical Chern-Simons action is normalized as

ik2π12(A,dA)\displaystyle\frac{ik}{2\pi}\frac{1}{2}(A,*dA)

which is off the canonical normalization by the factor of iK2π𝑖𝐾2𝜋\frac{iK}{2\pi}, the perturbative computation of the path integral will give the result (assuming that p𝑝p is odd)

Z𝑍\displaystyle Z =\displaystyle= (iK2π)12b0(A(0))Vol(HA(0))τ(A(0))12+=1p12(iK2π)12b0(A())Vol(HA())e2πiK2/pτ(A())12superscript𝑖𝐾2𝜋12subscript𝑏0superscript𝐴0Volsubscript𝐻superscript𝐴0𝜏superscriptsuperscript𝐴012superscriptsubscript1𝑝12superscript𝑖𝐾2𝜋12subscript𝑏0superscript𝐴Volsubscript𝐻superscript𝐴superscript𝑒2𝜋𝑖𝐾superscript2𝑝𝜏superscriptsuperscript𝐴12\displaystyle\frac{\left(\frac{iK}{2\pi}\right)^{-\frac{1}{2}b_{0}(A^{(0)})}}{{\mbox{Vol}}\left(H_{A^{(0)}}\right)}\tau(A^{(0)})^{\frac{1}{2}}+\sum_{\ell=1}^{\frac{p-1}{2}}\frac{\left(\frac{iK}{2\pi}\right)^{-\frac{1}{2}b_{0}(A^{(\ell)})}}{{\mbox{Vol}}\left(H_{A^{(\ell)}}\right)}e^{2\pi iK\ell^{2}/p}\tau(A^{(\ell)})^{\frac{1}{2}}

We restrict the sum to run over =1,,(p1)/21𝑝12\ell=1,...,(p-1)/2 following Eq (2.18) in [7], Eq (4.17) in [9], and [8]. Here the RS torsions are given by

τ(A(0))𝜏superscript𝐴0\displaystyle\tau(A^{(0)}) =\displaystyle= (1p)3superscript1𝑝3\displaystyle\left(\frac{1}{p}\right)^{3}

the power 3 because there are three generators of SU(2)𝑆𝑈2SU(2), and

τ(A())𝜏superscript𝐴\displaystyle\tau(A^{(\ell)}) =\displaystyle= 1p(2sin2πp)41𝑝superscript22𝜋𝑝4\displaystyle\frac{1}{p}\left(2\sin\frac{2\pi\ell}{p}\right)^{4}

The volume of SU(2)=S3𝑆𝑈2superscript𝑆3SU(2)=S^{3} with unit radius r𝑟r is Vol(SU(2))=2π2r3Vol𝑆𝑈22superscript𝜋2superscript𝑟3{\mbox{Vol}}(SU(2))=2\pi^{2}r^{3} and the length of the equator is Vol(U(1))=2πrVol𝑈12𝜋𝑟{\mbox{Vol}}(U(1))=2\pi r.

Using b0(A(0))=3subscript𝑏0superscript𝐴03b_{0}(A^{(0)})=3 and b0(A())=1subscript𝑏0superscript𝐴1b_{0}(A^{({{\ell}})})=1, and by comparing with the known exact result to be presented in below, we get

Vol(SU(2))Vol𝑆𝑈2\displaystyle{\mbox{Vol}}(SU(2)) =\displaystyle= 2π2𝜋\displaystyle 2\sqrt{\pi}
Vol(U(1))Vol𝑈1\displaystyle{\mbox{Vol}}(U(1)) =\displaystyle= 2π2𝜋\displaystyle 2\sqrt{\pi}

This is consistent with taking the radius of SU(2)𝑆𝑈2SU(2) as

r𝑟\displaystyle r =\displaystyle= 1π1𝜋\displaystyle\frac{1}{\sqrt{\pi}}

E.2 The exact result

For SU(2)𝑆𝑈2SU(2) gauge group, the exact result for the partition function on L(p;1)𝐿𝑝1L(p;1) is given by

Z(ϵ)𝑍italic-ϵ\displaystyle Z(\epsilon) =\displaystyle= e3πi4i(3p)ϵ4=0p1C()dz2πisinh2(z2)eipz24ϵ2πϵzsuperscript𝑒3𝜋𝑖4𝑖3𝑝italic-ϵ4superscriptsubscript0𝑝1subscriptsuperscript𝐶𝑑𝑧2𝜋𝑖superscript2𝑧2superscript𝑒𝑖𝑝superscript𝑧24italic-ϵ2𝜋italic-ϵ𝑧\displaystyle-e^{\frac{3\pi i}{4}-\frac{i(3-p)\epsilon}{4}}\sum_{{{\ell}}=0}^{p-1}\int_{C^{({{\ell}})}}\frac{dz}{2\pi i}\sinh^{2}\left(\frac{z}{2}\right)e^{\frac{ipz^{2}}{4\epsilon}-\frac{2\pi{{\ell}}}{\epsilon}z}

where C()superscript𝐶C^{({{\ell}})} is the contour

z𝑧\displaystyle z =\displaystyle= eiπ4x4πipsuperscript𝑒𝑖𝜋4𝑥4𝜋𝑖𝑝\displaystyle e^{\frac{i\pi}{4}}x-\frac{4\pi i{{\ell}}}{p}

for =0,,p10𝑝1{{\ell}}=0,...,p-1. (This result can be extracted from Eq (5.38) in [2] by taking P=1𝑃1P=1, N=0𝑁0N=0, d=p𝑑𝑝d=p and θ0=3psubscript𝜃03𝑝\theta_{0}=3-p in the expression there.) Here

ϵitalic-ϵ\displaystyle\epsilon =\displaystyle= 2πk+22𝜋𝑘2\displaystyle\frac{2\pi}{k+2}

where k+2𝑘2k+2 is the shifted Chern-Simons level.

For p=1𝑝1p=1 the formula reproduces the famous result [1]

Z(S3)𝑍superscript𝑆3\displaystyle Z(S^{3}) =\displaystyle= 2k+2sin(πk+2)2𝑘2𝜋𝑘2\displaystyle\sqrt{\frac{2}{k+2}}\sin\left(\frac{\pi}{k+2}\right)

for the partition function on S3superscript𝑆3S^{3}.

For generic p𝑝p, the integrals can also be computed exactly with the following result

Z(ϵ)𝑍italic-ϵ\displaystyle Z(\epsilon) =\displaystyle= 12iϵπpeπi4(p3)=0p1e4π2iϵ2p(eiϵpcos4πp1)12𝑖italic-ϵ𝜋𝑝superscript𝑒𝜋𝑖4𝑝3superscriptsubscript0𝑝1superscript𝑒4superscript𝜋2𝑖italic-ϵsuperscript2𝑝superscript𝑒𝑖italic-ϵ𝑝4𝜋𝑝1\displaystyle\frac{1}{2i}\sqrt{\frac{\epsilon}{\pi p}}e^{\frac{\pi i}{4}(p-3)}\sum_{{{\ell}}=0}^{p-1}e^{\frac{4\pi^{2}i}{\epsilon}\frac{{{\ell}}^{2}}{p}}\left(e^{\frac{i\epsilon}{p}}\cos\frac{4\pi{{\ell}}}{p}-1\right)

To also see the shift from k𝑘k to k+2𝑘2k+2 we would need to compute the eta invariant.

Because the =00{{\ell}}=0 term has different leading term K𝐾K asymptotics from the terms with >00{{\ell}}>0, we separate the sum into these two pieces and pick up only the leading term from each piece

Z0(ϵ)subscript𝑍0italic-ϵ\displaystyle Z_{0}(\epsilon) =\displaystyle= 12iϵπpeπi4(p3)(eiϵp1)12𝑖italic-ϵ𝜋𝑝superscript𝑒𝜋𝑖4𝑝3superscript𝑒𝑖italic-ϵ𝑝1\displaystyle\frac{1}{2i}\sqrt{\frac{\epsilon}{\pi p}}e^{\frac{\pi i}{4}(p-3)}\left(e^{\frac{i\epsilon}{p}}-1\right)
=\displaystyle= eπi4p2π(1iKp)3/2superscript𝑒𝜋𝑖4𝑝2𝜋superscript1𝑖𝐾𝑝32\displaystyle e^{\frac{\pi i}{4}p}\sqrt{2}\pi\left(\frac{1}{iKp}\right)^{3/2}

and

Zrest(ϵ)subscript𝑍𝑟𝑒𝑠𝑡italic-ϵ\displaystyle Z_{rest}(\epsilon) =\displaystyle= 12iϵπpeπi4(p3)=1p1e4π2iϵ2p(cos4πp1)12𝑖italic-ϵ𝜋𝑝superscript𝑒𝜋𝑖4𝑝3superscriptsubscript1𝑝1superscript𝑒4superscript𝜋2𝑖italic-ϵsuperscript2𝑝4𝜋𝑝1\displaystyle\frac{1}{2i}\sqrt{\frac{\epsilon}{\pi p}}e^{\frac{\pi i}{4}(p-3)}\sum_{{{\ell}}=1}^{p-1}e^{\frac{4\pi^{2}i}{\epsilon}\frac{{{\ell}}^{2}}{p}}\left(\cos\frac{4\pi{{\ell}}}{p}-1\right)
=\displaystyle= eπi4p212iKp=1p1e2πiK2p(sin2πp)2superscript𝑒𝜋𝑖4𝑝212𝑖𝐾𝑝superscriptsubscript1𝑝1superscript𝑒2𝜋𝑖𝐾superscript2𝑝superscript2𝜋𝑝2\displaystyle e^{\frac{\pi i}{4}p}2\sqrt{\frac{1}{2iKp}}\sum_{{{\ell}}=1}^{p-1}e^{\frac{2\pi iK{{\ell}}^{2}}{p}}\left(\sin\frac{2\pi{{\ell}}}{p}\right)^{2}

There will be an order K3/2superscript𝐾32K^{-3/2} contribution to Zrestsubscript𝑍𝑟𝑒𝑠𝑡Z_{rest} as well, but we can ignore that since each {{\ell}}-sector can be studied on its own. By a complex conjugation ii𝑖𝑖i\rightarrow-i, we have now obtained Eq (2.37) in [8].

The result in [8] was presented for odd p𝑝p. In that case, the sum can be replaced by twice of half of the sum as

=1p1e2πiK2p(sin2πp)2=2=1p12e2πiK2p(sin2πp)2superscriptsubscript1𝑝1superscript𝑒2𝜋𝑖𝐾superscript2𝑝superscript2𝜋𝑝22superscriptsubscript1𝑝12superscript𝑒2𝜋𝑖𝐾superscript2𝑝superscript2𝜋𝑝2\displaystyle\sum_{{{\ell}}=1}^{p-1}e^{\frac{2\pi iK{{\ell}}^{2}}{p}}\left(\sin\frac{2\pi{{\ell}}}{p}\right)^{2}=2\sum_{{{\ell}}=1}^{\frac{p-1}{2}}e^{\frac{2\pi iK{{\ell}}^{2}}{p}}\left(\sin\frac{2\pi{{\ell}}}{p}\right)^{2}

On the other hand, if p𝑝p is even, then the RS torsion is vanishing for =p/2𝑝2\ell=p/2 and we can write

=1p1e2πiK2p(sin2πp)2=20<<p/2e2πiK2p(sin2πp)2superscriptsubscript1𝑝1superscript𝑒2𝜋𝑖𝐾superscript2𝑝superscript2𝜋𝑝22subscript0𝑝2superscript𝑒2𝜋𝑖𝐾superscript2𝑝superscript2𝜋𝑝2\displaystyle\sum_{{{\ell}}=1}^{p-1}e^{\frac{2\pi iK{{\ell}}^{2}}{p}}\left(\sin\frac{2\pi{{\ell}}}{p}\right)^{2}=2\sum_{0<{{\ell}}<p/2}e^{\frac{2\pi iK{{\ell}}^{2}}{p}}\left(\sin\frac{2\pi{{\ell}}}{p}\right)^{2}

in agreement with [7].

Appendix F Dimensional reduction of selfdual forms on a circle

We consider a nonselfdual 2k2𝑘2k-form potential on Euclidean S1×M4k+1superscript𝑆1subscript𝑀4𝑘1S^{1}\times M_{4k+1}. The ghost hierarchi grows linearly, which gives the partition functions as

Z4k+2subscript𝑍4𝑘2\displaystyle Z_{4k+2} =\displaystyle= =1k(det2k2+1)22=0k(det2k2)2+12superscriptsubscriptproduct1𝑘superscriptsuperscriptsubscript2𝑘2122superscriptsubscriptproduct0𝑘superscriptsuperscriptsubscript2𝑘2212\displaystyle\frac{\prod_{{{\ell}}=1}^{k}\left(\det{}^{\prime}\triangle_{2k-2{{\ell}}+1}\right)^{\frac{2{{\ell}}}{2}}}{\prod_{{{\ell}}=0}^{k}\left(\det{}^{\prime}\triangle_{2k-2{{\ell}}}\right)^{\frac{2{{\ell}}+1}{2}}}

The partition function of a (2k1)2𝑘1(2k-1)-potential on M4k+1subscript𝑀4𝑘1M_{4k+1} is likewise given by

Z4k+1subscript𝑍4𝑘1\displaystyle Z_{4k+1} =\displaystyle= =1k(det2k2)22=1k(det2k2+1)212superscriptsubscriptproduct1𝑘superscriptsuperscriptsubscript2𝑘222superscriptsubscriptproduct1𝑘superscriptsuperscriptsubscript2𝑘21212\displaystyle\frac{\prod_{{{\ell}}=1}^{k}\left(\det{}^{\prime}\triangle_{2k-2{{\ell}}}\right)^{\frac{2{{\ell}}}{2}}}{\prod_{{{\ell}}=1}^{k}\left(\det{}^{\prime}\triangle_{2k-2{{\ell}}+1}\right)^{\frac{2{{\ell}}-1}{2}}}

We then dimensionally reduce along S1superscript𝑆1S^{1} by replacing detpsuperscriptsubscript𝑝\det{}^{\prime}\triangle_{p} by detpdetp1superscriptsubscript𝑝superscriptsubscript𝑝1\det{}^{\prime}\triangle_{p}\det{}^{\prime}\triangle_{p-1} where the latter represent Laplacians on M4k+1subscript𝑀4𝑘1M_{4k+1}. This gives the dimensionally reduced partition function as

Z4k+2(τ=0)subscript𝑍4𝑘2subscript𝜏0\displaystyle Z_{4k+2}(\partial_{\tau}=0) =\displaystyle= =1k(det2k2+1)12=0k(det2k2)12superscriptsubscriptproduct1𝑘superscriptsuperscriptsubscript2𝑘2112superscriptsubscriptproduct0𝑘superscriptsuperscriptsubscript2𝑘212\displaystyle\frac{\prod_{{{\ell}}=1}^{k}\left(\det{}^{\prime}\triangle_{2k-2{{\ell}}+1}\right)^{\frac{1}{2}}}{\prod_{{{\ell}}=0}^{k}\left(\det{}^{\prime}\triangle_{2k-2{{\ell}}}\right)^{\frac{1}{2}}}

We then compute the ratio

Z4k+2(τ=0)Z4k+12subscript𝑍4𝑘2subscript𝜏0superscriptsubscript𝑍4𝑘12\displaystyle\frac{Z_{4k+2}(\partial_{\tau}=0)}{Z_{4k+1}^{2}} =\displaystyle= =1k(det2k2+1)212=0k(det2k2)2+12superscriptsubscriptproduct1𝑘superscriptsuperscriptsubscript2𝑘21212superscriptsubscriptproduct0𝑘superscriptsuperscriptsubscript2𝑘2212\displaystyle\frac{\prod_{{{\ell}}=1}^{k}\left(\det{}^{\prime}\triangle_{2k-2{{\ell}}+1}\right)^{2{{\ell}}-\frac{1}{2}}}{\prod_{{{\ell}}=0}^{k}\left(\det{}^{\prime}\triangle_{2k-2{{\ell}}}\right)^{2{{\ell}}+\frac{1}{2}}}

Using Poincare duality, we get the Ray-Singer torsion on a (4k+1)4𝑘1(4k+1)-dimensional manifold M4k+1subscript𝑀4𝑘1M_{4k+1} as

τosc(M4k+1)subscript𝜏𝑜𝑠𝑐subscript𝑀4𝑘1\displaystyle\tau_{osc}(M_{4k+1}) =\displaystyle= p=02k(detp)(1)p4k+12p2superscriptsubscriptproduct𝑝02𝑘superscriptsuperscriptsubscript𝑝superscript1𝑝4𝑘12𝑝2\displaystyle\prod_{p=0}^{2k}\left(\det{}^{\prime}\triangle_{p}\right)^{(-1)^{p}\frac{4k+1-2p}{2}}
=\displaystyle= =0k(det2k2)2+12=1k(det2k2+1)212superscriptsubscriptproduct0𝑘superscriptsuperscriptsubscript2𝑘2212superscriptsubscriptproduct1𝑘superscriptsuperscriptsubscript2𝑘21212\displaystyle\frac{\prod_{{{\ell}}=0}^{k}\left(\det{}^{\prime}\triangle_{2k-2{{\ell}}}\right)^{2{{\ell}}+\frac{1}{2}}}{\prod_{{{\ell}}=1}^{k}\left(\det{}^{\prime}\triangle_{2k-2{{\ell}}+1}\right)^{2{{\ell}}-\frac{1}{2}}}

and we see that

Z4k+2(τ=0)Z4k+12subscript𝑍4𝑘2subscript𝜏0superscriptsubscript𝑍4𝑘12\displaystyle\frac{Z_{4k+2}(\partial_{\tau}=0)}{Z_{4k+1}^{2}} =\displaystyle= 1τosc(M4k+1)1subscript𝜏𝑜𝑠𝑐subscript𝑀4𝑘1\displaystyle\frac{1}{\tau_{osc}(M_{4k+1})}

holds for any k=0,1,2,3,𝑘0123k=0,1,2,3,.... For the case k=0𝑘0k=0 which corresponds to a zero-form in 2d, the relation still holds if we assume that the 1d oscillator partition function is equal to one.

In 4k14𝑘14k-1 dimensions, the analytic torsion is

τoscsubscript𝜏𝑜𝑠𝑐\displaystyle\tau_{osc} =\displaystyle= =1k(det2k2)212=0k1(det2k21)2+12superscriptsubscriptproduct1𝑘superscriptsuperscriptsubscript2𝑘2212superscriptsubscriptproduct0𝑘1superscriptsuperscriptsubscript2𝑘21212\displaystyle\frac{\prod_{{{\ell}}=1}^{k}\left(\det{}^{\prime}\triangle_{2k-2{{\ell}}}\right)^{{2{{\ell}}-\frac{1}{2}}}}{\prod_{{{\ell}}=0}^{k-1}\left(\det{}^{\prime}\triangle_{2k-2{{\ell}}-1}\right)^{{2{{\ell}}+\frac{1}{2}}}}

We have

Z4ksubscript𝑍4𝑘\displaystyle Z_{4k} =\displaystyle= =1k(det2k2)22=0k1(det2k21)2+12superscriptsubscriptproduct1𝑘superscriptsuperscriptsubscript2𝑘222superscriptsubscriptproduct0𝑘1superscriptsuperscriptsubscript2𝑘21212\displaystyle\frac{\prod_{{{\ell}}=1}^{k}\left(\det{}^{\prime}\triangle_{2k-2{{\ell}}}\right)^{\frac{2{{\ell}}}{2}}}{\prod_{{{\ell}}=0}^{k-1}\left(\det{}^{\prime}\triangle_{2k-2{{\ell}}-1}\right)^{\frac{2{{\ell}}+1}{2}}}

and

Z4k1subscript𝑍4𝑘1\displaystyle Z_{4k-1} =\displaystyle= =0k1(det2k21)22=0k1(det2k(2+2))2+12superscriptsubscriptproduct0𝑘1superscriptsuperscriptsubscript2𝑘2122superscriptsubscriptproduct0𝑘1superscriptsuperscriptsubscript2𝑘22212\displaystyle\frac{\prod_{{{\ell}}=0}^{k-1}\left(\det{}^{\prime}\triangle_{2k-2{{\ell}}-1}\right)^{\frac{2{{\ell}}}{2}}}{\prod_{{{\ell}}=0}^{k-1}\left(\det{}^{\prime}\triangle_{2k-(2{{\ell}}+2)}\right)^{\frac{2{{\ell}}+1}{2}}}

Then

Z4k(τ=0)subscript𝑍4𝑘subscript𝜏0\displaystyle Z_{4k}(\partial_{\tau}=0) =\displaystyle= =1k(det2k2)12=0k1(det2k2=1)12superscriptsubscriptproduct1𝑘superscriptsuperscriptsubscript2𝑘212superscriptsubscriptproduct0𝑘1superscriptsuperscriptsubscript2𝑘2112\displaystyle\frac{\prod_{{{\ell}}=1}^{k}\left(\det{}^{\prime}\triangle_{2k-2{{\ell}}}\right)^{\frac{1}{2}}}{\prod_{{{\ell}}=0}^{k-1}\left(\det{}^{\prime}\triangle_{2k-2{{\ell}}=1}\right)^{\frac{1}{2}}}

Then the relation instead becomes

Z4k(τ=0)Z4k12subscript𝑍4𝑘subscript𝜏0superscriptsubscript𝑍4𝑘12\displaystyle\frac{Z_{4k}(\partial_{\tau}=0)}{Z_{4k-1}^{2}} =\displaystyle= τosc(M4k)subscript𝜏𝑜𝑠𝑐subscript𝑀4𝑘\displaystyle\tau_{osc}(M_{4k})

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